Kirchhoff's laws — practice questions

1. Identify Junctions and Label Currents: Let the current leaving the battery be I_total. This current splits into I1 through R1 and I2 through R2 at the first junction. They recombine to form I_total again before going through R3.
2. Apply Kirchhoff's First Law (KCL): At the junction where I_total splits: I_total = I1 + I2.
3. Identify Closed Loops and Apply Kirchhoff's Second Law (KVL):
   • Loop 1 (Battery -> R1 -> R3 -> Battery): (Assume clockwise traversal) +6V - (I1 * R1) - (I_total * R3) = 0 +6 - 3I1 - 2I_total = 0 (Equation 1)
   • Loop 2 (Battery -> R2 -> R3 -> Battery): (Assume clockwise traversal) +6V - (I2 * R2) - (I_total * R3) = 0 +6 - 6I2 - 2I_total = 0 (Equation 2)
   • Loop 3 (R1 -> R2 -> R1 -- internal loop): (Starting from top of R1, going clockwise) (I1 * R1) - (I2 * R2) = 0 3I1 - 6I2 = 0 3I1 = 6I2 $\Rightarrow I1 = 2*I2 (Equation 3)$$
4. Solve the System of Equations:
   • From KCL: I_total = I1 + I2
   • Substitute I1 = 2I2 into KCL: I_total = 2I2 + I2 = 3*I2.
   • Substitute I1 = 2I2 and I_total = 3I2 into Equation 1: 6 - 3*(2I2) - 2(3I2) = 0 6 - 6I2 - 6I2 = 0 6 - 12I2 = 0 12*I2 = 6 I2 = 0.5 A
   • Now find I1: I1 = 2*I2 = 2 * 0.5 = 1 A
   • Finally, find I_total: I_total = I1 + I2 = 1 + 0.5 = 1.5 A

Answer: The total current drawn from the battery is 1.5 A.

1. Draw the Circuit and Assume Current Direction: Let's assume the current $I$ flows clockwise, driven by the larger e.m.f. of Cell A. The current will flow out of the positive terminal of Cell A, through the 5.0 Ω resistor, and into the positive terminal of Cell B.
2. Apply Kirchhoff's Second Law (KVL): We will trace the loop in a clockwise direction, starting from the negative terminal of Cell A.
   • Move across Cell A (- to +): Potential rise = +9.0 V
   • Move across internal resistance of A (with current): Potential drop = - I * r_A = - I * 1.0
   • Move across the external resistor (with current): Potential drop = - I * R = - I * 5.0
   • Move across Cell B (+ to -): Potential drop = -3.0 V
   • Move across internal resistance of B (with current): Potential drop = - I * r_B = - I * 0.5
3. Form the KVL Equation: Sum of all potential changes is zero. $+9.0 - I(1.0) - I(5.0) - 3.0 - I(0.5) = 0$
4. Solve for the Current (I):
   • Group the e.m.f. terms and the resistance terms: $(9.0 - 3.0) - I(1.0 + 5.0 + 0.5) = 0$
   • Simplify: $6.0 - I(6.5) = 0$
   • Rearrange to find I: $6.0 = 6.5 * I$ $I = \frac{6.0}{6.5} \approx 0.92307... \text{ A}$
   • The current in the circuit is 0.92 A (to 2 s.f.). Since the value is positive, our assumed clockwise direction was correct.
5. Calculate Terminal p.d. across Cell A:
   • Cell A is discharging (supplying energy), so we use the formula V = ε - Ir.
   • $V_A = \epsilon_A - I * r_A$
   • $V_A = 9.0 - (0.923 * 1.0)$
   • $V_A = 9.0 - 0.923 = 8.077 \text{ V}$
   • The terminal potential difference across Cell A is 8.1 V (to 2 s.f.).

Answer: The current in the circuit is 0.92 A and the terminal p.d. across Cell A is 8.1 V.