Current, Resistance, and DC Circuits

Current and Resistance

Part 1 of 7 — Fundamentals of DC Circuits

Electric Current

Current is the rate at which charge flows through a cross-section of a conductor. As a calculus-based definition:

$I = \frac{dQ}{dt}$

If the current is not constant, the total charge that passes a point between times $t_1$ and $t_2$ is the integral of the current:

$Q = \int_{t_1}^{t_2} I\,dt$

Units: Amperes (A) = Coulombs/second.

Microscopic Picture: Drift Velocity

The macroscopic current is tied to the motion of mobile charge carriers (density $n$, charge $q$, drift speed $v_d$) in a wire of cross-sectional area $A$:

$I = n\,q\,v_d\,A$

Ohm's Law

$V = IR$

Resistivity

$R = \rho \frac{L}{A}$

where $\rho$ is resistivity, $L$ is length, $A$ is cross-sectional area. The microscopic form of Ohm's law relates the field to the current density $\vec{J}$: $\vec{E} = \rho\,\vec{J}$.

Power

$P = IV = I^2R = \frac{V^2}{R}$

Current (conventional) flows from high potential to low potential.

Worked Example — Charge from a Time-Varying Current

Problem. The current in a wire varies with time as $I(t) = (3.0\,\text{A/s}^2)\,t^2 + (2.0\,\text{A})$. How much charge flows past a point during the first $4.0$ seconds, and what constant current would transport the same charge in that interval?

Step 1 — Set up the integral. Because $I = dQ/dt$, the charge is

$Q = \int_0^{4} I(t)\,dt = \int_0^{4}\left(3t^2 + 2\right)dt$

Step 2 — Integrate term by term.

$Q = \left[\,t^3 + 2t\,\right]_0^{4}$

Step 3 — Evaluate the bounds.

$Q = (4^3 + 2\cdot 4) - 0 = 64 + 8 = 72\,\text{C}$

Step 4 — Equivalent constant current. The average current is the total charge divided by the elapsed time:

$I_{\text{avg}} = \frac{Q}{\Delta t} = \frac{72\,\text{C}}{4.0\,\text{s}} = 18\,\text{A}$

Takeaway. When current depends on time, you integrate $I(t)$ to get charge; a single "plug-in" of $I \times t$ only works for constant current.

Resistance vs. Resistivity — Don't Confuse Them

A common AP trap is treating resistivity $\rho$ and resistance $R$ as interchangeable.

• Resistivity $\rho$ is an intrinsic material property (units $\Omega\cdot\text{m}$). Copper has $\rho \approx 1.7\times10^{-8}\,\Omega\cdot\text{m}$ no matter how the wire is shaped.
• Resistance $R = \rho L/A$ depends on geometry — stretch the wire or change its cross-section and $R$ changes even though $\rho$ does not.

Temperature dependence. For many conductors, resistivity rises roughly linearly with temperature:

$\rho(T) = \rho_0\left[1 + \alpha(T - T_0)\right]$

where $\alpha$ is the temperature coefficient of resistivity. This is why a light-bulb filament has a much larger resistance when hot than when cold.

Think "$\rho$ = the material, $R$ = the material plus the shape."

Concept Check

Series and Parallel Circuits

Part 2 of 7 — Combining Resistors

Series Resistors

$R_{\text{eq}} = R_1 + R_2 + R_3 + \cdots$

• Same current through each resistor
• Voltages add: $V = V_1 + V_2 + V_3$
• Equivalent resistance is always larger than the largest resistor

Parallel Resistors

$\frac{1}{R_{\text{eq}}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + \cdots$

For two resistors: $R_{\text{eq}} = \frac{R_1 R_2}{R_1 + R_2}$

• Same voltage across each resistor
• Currents add: $I = I_1 + I_2 + I_3$
• Equivalent resistance is always smaller than the smallest resistor

Power in Combinations

In series the largest resistor dissipates the most power ($P = I^2R$ with shared $I$); in parallel the smallest resistor dissipates the most ($P = V^2/R$ with shared $V$).

Series: current same, voltage divides. Parallel: voltage same, current divides.

Worked Example — Reducing a Mixed Network

Problem. A $12\,\text{V}$ battery (no internal resistance) is connected to a $4\,\Omega$ resistor in series with the parallel combination of a $6\,\Omega$ and a $3\,\Omega$ resistor. Find (a) the total current from the battery and (b) the power dissipated in the $6\,\Omega$ resistor.

Step 1 — Combine the parallel pair.

$R_{\text{par}} = \frac{R_1 R_2}{R_1 + R_2} = \frac{(6)(3)}{6 + 3} = \frac{18}{9} = 2\,\Omega$

Step 2 — Add the series resistor.

$R_{\text{eq}} = 4 + 2 = 6\,\Omega$

Step 3 — Total current (Ohm's law on the whole circuit).

$I_{\text{total}} = \frac{V}{R_{\text{eq}}} = \frac{12}{6} = 2\,\text{A}$

Step 4 — Voltage across the parallel section. That same $2\,\text{A}$ flows through the $4\,\Omega$ and then through the parallel block:

$V_{\text{par}} = I_{\text{total}}\,R_{\text{par}} = (2)(2) = 4\,\text{V}$

Step 5 — Power in the $6\,\Omega$ resistor. It sees $4\,\text{V}$ across it:

$P_{6} = \frac{V_{\text{par}}^2}{R} = \frac{(4)^2}{6} = \frac{16}{6} \approx 2.7\,\text{W}$

Takeaway. Collapse parallel blocks first, ride the series current to find branch voltages, then use $P = V^2/R$ on the individual resistor.

Worked Example (Calculus) — Resistance of a Tapered Conductor

Problem. A conductor of resistivity $\rho$ and length $L$ is shaped so its circular cross-section grows linearly: the radius is $a$ at one end and $b$ at the other. Find its total resistance by treating it as infinitely many thin disks in series.

Step 1 — Slice it. A disk of thickness $dx$ at position $x$ (from the small end) behaves like a tiny series resistor:

$dR = \frac{\rho\,dx}{A(x)} = \frac{\rho\,dx}{\pi r(x)^2}$

Step 2 — Express the radius as a function of $x$. Linear taper from $a$ to $b$ over length $L$:

$r(x) = a + \frac{(b - a)}{L}\,x$

Step 3 — Sum the series of slices by integrating.

$R = \int_0^{L}\frac{\rho\,dx}{\pi\,r(x)^2} = \frac{\rho}{\pi}\int_0^{L}\frac{dx}{\left(a + \frac{b-a}{L}x\right)^2}$

Step 4 — Substitute $u = a + \frac{b-a}{L}x$, so $du = \frac{b-a}{L}\,dx$:

$R = \frac{\rho}{\pi}\cdot\frac{L}{b-a}\int_{a}^{b}\frac{du}{u^2} = \frac{\rho L}{\pi(b-a)}\left[-\frac{1}{u}\right]_{a}^{b}$

Step 5 — Evaluate.

$R = \frac{\rho L}{\pi(b-a)}\left(\frac{1}{a} - \frac{1}{b}\right) = \frac{\rho L}{\pi(b-a)}\cdot\frac{b-a}{ab} = \frac{\rho L}{\pi a b}$

Check. If $a = b = r$ (a uniform cylinder), this reduces to $R = \rho L/(\pi r^2) = \rho L/A$, the familiar formula.

Takeaway. When the cross-section varies, slice the conductor into series disks and integrate $dR = \rho\,dx/A(x)$ — series combination becomes a definite integral.

Voltage Dividers and Current Dividers

Two shortcuts that save time on exams:

Voltage divider (series). When resistors $R_1$ and $R_2$ are in series across a source voltage $V$, the voltage across $R_1$ is its share of the total resistance:

$V_1 = V\,\frac{R_1}{R_1 + R_2}$

Current divider (parallel). When current $I$ splits between $R_1$ and $R_2$ in parallel, the branch with the smaller resistance gets the larger share. For two resistors:

$I_1 = I\,\frac{R_2}{R_1 + R_2}$

Notice the "opposite" resistor appears in the numerator of the current divider — current prefers the path of least resistance.

Series divides voltage in direct proportion to resistance; parallel divides current in inverse proportion to resistance.

Concept Check

Kirchhoff's Rules

Part 3 of 7 — Analyzing Complex Circuits

Junction Rule (KCL)

$\sum I_{\text{in}} = \sum I_{\text{out}}$

This is a statement of conservation of charge at any junction (node).

Loop Rule (KVL)

$\sum \Delta V = 0$

Around any closed loop, the total change in potential is zero — a statement of energy conservation.

Sign Conventions:

Write enough independent equations (junction + loop) to solve for all unknown currents.

Worked Example — A Two-Loop Circuit

Problem. A battery $\mathcal{E} = 12\,\text{V}$ connects to node A. From A, two parallel branches run to node B: branch 1 has $R_1 = 4\,\Omega$, branch 2 has $R_2 = 6\,\Omega$. From B the current returns through $R_3 = 2\,\Omega$ back to the battery. Find the current $I_3$ through $R_3$.

Step 1 — Recognize structure. $R_1 \parallel R_2$, then in series with $R_3$. Let $I_1$, $I_2$ be the branch currents and $I_3$ the return current.

Step 2 — Junction rule at A.

$I_3 = I_1 + I_2$

Step 3 — Both branches share the same A-to-B voltage $V_{AB}$, so $I_1 = V_{AB}/4$ and $I_2 = V_{AB}/6$. Substituting:

$I_3 = \frac{V_{AB}}{4} + \frac{V_{AB}}{6} = V_{AB}\left(\frac{3 + 2}{12}\right) = \frac{5V_{AB}}{12}$

Step 4 — Loop rule (battery, through B, through $R_3$).

$\mathcal{E} - V_{AB} - I_3 R_3 = 0 \;\Rightarrow\; 12 = V_{AB} + 2I_3$

Step 5 — Solve the system. From Step 3, $V_{AB} = \tfrac{12}{5}I_3 = 2.4\,I_3$. Substitute:

$12 = 2.4\,I_3 + 2I_3 = 4.4\,I_3 \;\Rightarrow\; I_3 = \frac{12}{4.4} \approx 2.7\,\text{A}$

Check. $R_1 \parallel R_2 = (4)(6)/10 = 2.4\,\Omega$, so $R_{eq} = 2.4 + 2 = 4.4\,\Omega$ and $I_3 = 12/4.4 \approx 2.7\,\text{A}$. Consistent.

Worked Example (Calculus) — The Loop Rule Produces a Differential Equation

The loop rule is not limited to constant currents. Applied to a circuit with a capacitor, it becomes a differential equation in time. Consider a battery $\mathcal{E}$, resistor $R$, and initially uncharged capacitor $C$ in a single charging loop.

Step 1 — Write the loop rule, term by term. Traveling around the loop (battery rise, resistor drop, capacitor drop):

$\mathcal{E} - IR - \frac{q}{C} = 0$

Step 2 — Replace $I$ with its calculus definition. During charging the capacitor's charge grows, so $I = +\dfrac{dq}{dt}$:

$\mathcal{E} - R\frac{dq}{dt} - \frac{q}{C} = 0$

Step 3 — Rearrange into a first-order ODE.

$R\frac{dq}{dt} = \mathcal{E} - \frac{q}{C} \;\Rightarrow\; \frac{dq}{dt} = \frac{1}{R}\left(\mathcal{E} - \frac{q}{C}\right)$

Step 4 — Separate variables and integrate. With $q(0) = 0$ and final charge $q$:

$\int_0^{q}\frac{dq'}{\,\mathcal{E} - q'/C\,} = \int_0^{t}\frac{dt'}{R}$

The left side integrates to $-C\ln\!\left(\mathcal{E} - q/C\right)$ evaluated from $0$ to $q$, giving

$-C\ln\!\frac{\mathcal{E} - q/C}{\mathcal{E}} = \frac{t}{R}$

Step 5 — Solve for $q(t)$. Exponentiate and rearrange:

$q(t) = C\mathcal{E}\left(1 - e^{-t/RC}\right)$

Takeaway. Kirchhoff's loop rule plus $I = dq/dt$ turns directly into a separable differential equation; the exponential charging law is its solution, not a separate formula to memorize.

A Reliable Recipe for Multi-Loop Circuits

When the network won't reduce by simple series/parallel collapsing (e.g. a Wheatstone-bridge layout), fall back on the full Kirchhoff procedure:

1. Assign a current (with an assumed direction) to every branch. If there are $b$ branches with unknown currents, you need $b$ independent equations.
2. Junction equations. With $n$ nodes, exactly $n - 1$ of the junction equations are independent.
3. Loop equations. Make up the remaining $b - (n - 1)$ equations from independent loops, applying the sign conventions consistently.
4. Solve the linear system (substitution or matrices).
5. Interpret signs. A negative current just means the true direction is opposite your guess.

Energy bookkeeping. Once all currents are known, the power delivered by each EMF, $P = \mathcal{E} I$, must equal the total dissipated, $\sum I^2 R$ — a built-in check on your algebra.

Count branches and nodes first; that tells you exactly how many equations to write — no more, no less.

Concept Check

RC Circuits

Part 4 of 7 — Charging and Discharging Capacitors

Charging an RC Circuit

$q(t) = C\mathcal{E}\left(1 - e^{-t/RC}\right)$

$I(t) = \frac{\mathcal{E}}{R}\,e^{-t/RC}$

$V_C(t) = \mathcal{E}\left(1 - e^{-t/RC}\right)$

Discharging an RC Circuit

$q(t) = Q_0\,e^{-t/RC}$

$I(t) = -\frac{Q_0}{RC}\,e^{-t/RC}$

Time Constant

$\tau = RC$

After about 5 time constants, the circuit is essentially at steady state.

Worked Example — Deriving the Discharge Curve from the Differential Equation

Problem. A capacitor $C$ initially holds charge $Q_0$. At $t=0$ it is connected across a resistor $R$ with no battery. Derive $q(t)$ and the current, then evaluate for $C = 2.0\,\mu\text{F}$, $R = 5.0\times 10^{5}\,\Omega$, $Q_0 = 8.0\,\mu\text{C}$ at $t = 1.0\,\text{s}$.

Step 1 — Apply the loop rule. The capacitor voltage $q/C$ drives the current through $R$:

$\frac{q}{C} - IR = 0, \qquad I = -\frac{dq}{dt}$

(The minus sign appears because the capacitor's charge decreases as current flows.)

Step 2 — Form the differential equation.

$\frac{q}{C} = -R\frac{dq}{dt} \;\Rightarrow\; \frac{dq}{dt} = -\frac{q}{RC}$

Step 3 — Separate variables and integrate.

$\int_{Q_0}^{q}\frac{dq'}{q'} = -\frac{1}{RC}\int_0^{t}dt' \;\Rightarrow\; \ln\!\frac{q}{Q_0} = -\frac{t}{RC}$

Step 4 — Exponentiate.

$q(t) = Q_0\,e^{-t/RC}$

Step 5 — Differentiate to get current.

$I(t) = -\frac{dq}{dt} = \frac{Q_0}{RC}\,e^{-t/RC}$

Step 6 — Plug in numbers. First the time constant:

$\tau = RC = (5.0\times10^{5})(2.0\times10^{-6}) = 1.0\,\text{s}$

At $t = 1.0\,\text{s}$ we have $t/\tau = 1$, so:

$q = (8.0\,\mu\text{C})\,e^{-1} = (8.0)(0.368) \approx 2.9\,\mu\text{C}$

Takeaway. Every RC result comes from the same move: loop rule gives a first-order separable ODE, and integration produces the exponential.

Reading the Exponential: Slopes, Half-Life, and Limits

The time constant is a slope, not just a clock. For discharge $q = Q_0 e^{-t/RC}$, the initial rate of change is

$\left.\frac{dq}{dt}\right|_{t=0} = -\frac{Q_0}{RC}$

If the capacitor kept discharging at that initial rate, it would reach zero in exactly one time constant $\tau = RC$. The actual curve bends, so it instead reaches $36.8\%$ at $t = \tau$.

Half-life. The time to fall to half the charge satisfies $\tfrac12 = e^{-t_{1/2}/RC}$, giving

$t_{1/2} = RC\ln 2 \approx 0.693\,RC$

The two limits worth memorizing:

• At $t = 0^+$: an uncharged capacitor acts like a wire (max current); a charged one acts like a battery.
• As $t \to \infty$: a fully charged capacitor acts like an open switch (no current).

Sketching the tangent line at $t=0$ and marking the $\tau$, $2\tau$, $5\tau$ gridlines turns any RC problem into a quick graph.

Concept Check

EMF and Internal Resistance

Part 5 of 7 — Real Batteries

Electromotive Force (EMF)

EMF ($\mathcal{E}$) is the potential difference a battery provides with no current flowing (open circuit).

With internal resistance $r$, the terminal voltage under load is:

$V_{\text{terminal}} = \mathcal{E} - Ir$

For a single external resistor $R$, the current is set by both resistances:

$I = \frac{\mathcal{E}}{R + r}$

Power Delivered

$P_{\text{delivered}} = I^2 R_{\text{external}}$

$P_{\text{wasted}} = I^2 r$

Maximum power transfer to the load occurs when $R_{\text{ext}} = r$.

Internal resistance means the terminal voltage drops as the load current increases.

Worked Example — Maximum Power Transfer (a Calculus Optimization)

Problem. A battery has EMF $\mathcal{E}$ and internal resistance $r$. Show that the external resistance $R$ which maximizes the power delivered to the load is $R = r$, and find that maximum power.

Step 1 — Write the load power as a function of $R$. The current is $I = \mathcal{E}/(R+r)$, so

$P(R) = I^2 R = \frac{\mathcal{E}^2 R}{(R + r)^2}$

Step 2 — Differentiate with respect to $R$ (quotient rule):

$\frac{dP}{dR} = \mathcal{E}^2\,\frac{(R+r)^2 - R\cdot 2(R+r)}{(R+r)^4}$

Step 3 — Simplify the numerator. Factor $(R+r)$:

$\frac{dP}{dR} = \mathcal{E}^2\,\frac{(R+r) - 2R}{(R+r)^3} = \mathcal{E}^2\,\frac{r - R}{(R+r)^3}$

Step 4 — Set the derivative to zero. The denominator is always positive, so

$r - R = 0 \;\Rightarrow\; R = r$

Because $dP/dR > 0$ for $R < r$ and $dP/dR < 0$ for $R > r$, this critical point is a maximum.

Step 5 — Evaluate the maximum power. Substitute $R = r$:

$P_{\max} = \frac{\mathcal{E}^2\,r}{(2r)^2} = \frac{\mathcal{E}^2}{4r}$

Takeaway. Impedance matching ($R = r$) is a genuine calculus optimization, and at the optimum exactly half the total power is delivered to the load (the other half is lost in $r$).

Efficiency vs. Maximum Power — A Subtle Trade-Off

Maximum power transfer is not the same as maximum efficiency.

Define efficiency as the fraction of the battery's chemical power that reaches the load:

$\eta = \frac{P_{\text{load}}}{P_{\text{total}}} = \frac{I^2 R}{I^2(R + r)} = \frac{R}{R + r}$

• At the maximum-power condition $R = r$, the efficiency is only $\eta = r/(2r) = 50\%$ — half the energy is wasted heating the battery's interior.
• Efficiency increases toward $100\%$ as $R \to \infty$ (large load), but then the amount of power delivered shrinks toward zero.

This is why power utilities use very low source resistance and high load resistance: they prioritize efficiency, not maximum power transfer.

"Most power" and "most efficient" pull in opposite directions; know which one a problem is asking for.

Concept Check

DC Circuits Workshop

Part 6 of 7 — Strategies

Circuit Analysis Steps

1. Simplify — combine series/parallel resistors where possible
2. Label — assign current directions and loop directions
3. Apply Kirchhoff's rules — write junction and loop equations
4. Solve — system of equations for the unknowns
5. Check — verify signs and units

For RC circuits: identify charging vs. discharging, then find $\tau = RC$. Remember the two limiting cases — at $t = 0^+$ an uncharged capacitor acts like a wire; as $t \to \infty$ a fully charged capacitor acts like an open switch.

Worked Example — Energy Stored and Heat Dissipated in Charging

Problem. A capacitor $C$ is charged from $0$ to full charge through a resistor $R$ by a battery $\mathcal{E}$. Using calculus, find (a) the total energy delivered by the battery, (b) the energy finally stored in the capacitor, and (c) the energy dissipated in $R$.

Step 1 — Battery energy. The battery pushes total charge $Q_f = C\mathcal{E}$ at constant EMF $\mathcal{E}$:

$W_{\text{batt}} = \mathcal{E}\,Q_f = \mathcal{E}(C\mathcal{E}) = C\mathcal{E}^2$

Step 2 — Energy stored in the capacitor.

$U_C = \frac{1}{2}\frac{Q_f^2}{C} = \frac{1}{2}C\mathcal{E}^2$

Step 3 — Energy dissipated in $R$ by direct integration. The charging current is $I(t) = (\mathcal{E}/R)e^{-t/RC}$, so the resistor dissipates

$W_R = \int_0^{\infty} I^2 R\,dt = \int_0^{\infty}\frac{\mathcal{E}^2}{R}\,e^{-2t/RC}\,dt$

Step 4 — Evaluate the integral. With $\int_0^\infty e^{-2t/RC}dt = RC/2$,

$W_R = \frac{\mathcal{E}^2}{R}\cdot\frac{RC}{2} = \frac{1}{2}C\mathcal{E}^2$

Step 5 — Energy balance. Indeed $W_{\text{batt}} = U_C + W_R$, since $C\mathcal{E}^2 = \tfrac12 C\mathcal{E}^2 + \tfrac12 C\mathcal{E}^2$.

Takeaway. Exactly half of the battery's energy ends up in the capacitor and half is dissipated as heat — independent of $R$. The integral of $I^2R$ is the rigorous way to see it.

Steady-State Analysis with Capacitors

Many AP problems show a resistor network with a capacitor and ask for the long-time (steady-state) behavior. The trick:

1. Replace each fully-charged capacitor with an open circuit. In DC steady state, $I = dq/dt = 0$ through the capacitor branch.
2. Solve the remaining purely-resistive circuit for currents and node voltages using series/parallel reduction or Kirchhoff.
3. The capacitor voltage equals the voltage across whatever it is connected in parallel with, computed from the resistive solution.
4. Stored charge then follows from $Q = CV_C$ and stored energy from $U = \tfrac12 C V_C^2$.

For the initial instant ($t = 0^+$) instead, replace an uncharged capacitor with a wire (short) and re-solve — that gives the maximum initial current.

Two snapshots — "capacitor as wire" at $t=0^+$ and "capacitor as open" at $t\to\infty$ — bracket the whole transient.

Concept Check

DC Circuits Review

Part 7 of 7 — Summary

Essential Formulas

Master the two RC limits ($t = 0^+$: capacitor = wire; $t \to \infty$: capacitor = open) and most exam problems fall out quickly.

Worked Example — A Capstone RC + Resistor Network

Problem. A $20\,\text{V}$ battery connects to a $10\,\Omega$ resistor in series with a parallel combination of a $20\,\Omega$ resistor and an (initially uncharged) capacitor $C = 5\,\mu\text{F}$. Find (a) the current right after the switch closes, (b) the steady-state voltage across the capacitor, and (c) the charging time constant.

Step 1 — At $t = 0^+$, the capacitor is a wire. It shorts out the parallel $20\,\Omega$ resistor, so the battery sees only the $10\,\Omega$:

$I_0 = \frac{\mathcal{E}}{R} = \frac{20}{10} = 2.0\,\text{A}$

Step 2 — At $t \to \infty$, the capacitor is an open switch. No current flows into the capacitor branch, so the current runs through $10\,\Omega$ and $20\,\Omega$ in series:

$I_\infty = \frac{20}{10 + 20} = \frac{20}{30} \approx 0.67\,\text{A}$

Step 3 — Steady-state capacitor voltage. The capacitor sits across the $20\,\Omega$ resistor, which carries $I_\infty$:

$V_C = I_\infty (20) = \left(\frac{2}{3}\right)(20) \approx 13.3\,\text{V}$

Step 4 — Time constant. For charging, the capacitor "sees" the Thévenin resistance: the $20\,\Omega$ in parallel with the $10\,\Omega$ (the battery is an ideal short for this purpose):

$R_{\text{Th}} = \frac{(10)(20)}{10 + 20} = \frac{200}{30} \approx 6.7\,\Omega$

$\tau = R_{\text{Th}}\,C = (6.7)(5\times10^{-6}) \approx 3.3\times10^{-5}\,\text{s}$

Takeaway. Use the $t=0^+$ and $t\to\infty$ limits for the endpoints, and the Thévenin resistance seen by the capacitor for $\tau$.

Units and Sanity Checks That Catch Errors

On a timed exam, a few quick checks save you from sign and factor mistakes:

Limiting cases are the most powerful check of all: let $R \to 0$, $R \to \infty$, $t \to 0$, or $t \to \infty$ and confirm your formula reduces to something obvious (a short, an open, the full EMF, or zero current).

If a symbolic answer fails a limiting-case check, the algebra is wrong — find the slip before plugging in numbers.

Concept Check