D.C. Circuits: Question 7

Worked solution

Part (a): Kirchhoff’s second law in each case

Kirchhoff’s second law states that, around the single loop, the e.m.f. equals the sum of the potential drops across the external resistor and the internal resistance: $E = IR + Ir = I(R+r)$

Applying this to each case: $\text{Case 1: } E = I_1(4.0+r) = 2.0(4.0+r)$ $\text{Case 2: } E = I_2(9.0+r) = 1.0(9.0+r)$

Part (b): Solving simultaneously for $E$ and $r$

Expanding both equations: $E = 8.0 + 2.0r \quad \text{...(1)}$ $E = 9.0 + 1.0r \quad \text{...(2)}$

Since both expressions equal $E$, set them equal to each other: $8.0 + 2.0r = 9.0 + 1.0r$ $2.0r - 1.0r = 9.0 - 8.0$ $r = 1.0\ \Omega$

Substituting back into equation (2): $E = 9.0 + 1.0(1.0) = 10.0\text{ V}$

Check using equation (1) instead: $E = 8.0+2.0(1.0)=8.0+2.0=10.0\text{ V}$, the same value. Both equations agree, confirming $E=10.0\text{ V}$ and $r=1.0\ \Omega$.

Part (c): Terminal potential difference in the first case

The terminal p.d. is the potential difference across the external resistor, found either directly or via $E-I_1r$: $V_{\text{terminal}} = I_1R_1 = 2.0\times4.0 = 8.0\text{ V}$

Check by the other route: $V_{\text{terminal}} = E - I_1r = 10.0 - (2.0\times1.0) = 10.0-2.0=8.0\text{ V}$. Both routes agree.

Part (d): Power dissipated in the internal resistance (second case)

Using the current $I_2=1.0\text{ A}$ that applies to the second case ($R=9.0\ \Omega$): $P_r = I_2^{\,2}r = (1.0)^2\times1.0 = 1.0\text{ W}$

Check using total and useful power: the total power supplied by the battery is $P_{\text{total}}=EI_2=10.0\times1.0=10.0\text{ W}$, and the useful power delivered to the $9.0\ \Omega$ resistor is $P_R=I_2^{\,2}R=(1.0)^2\times9.0=9.0\text{ W}$. Since $P_R+P_r=9.0+1.0=10.0\text{ W}=P_{\text{total}}$, energy is conserved and the answer is confirmed.

Final answers

• (a) $E=I_1(4.0+r)$ and $E=I_2(9.0+r)$
• (b) $E=\boxed{10.0}\text{ V}$, $r=\boxed{1.0}\ \Omega$
• (c) Terminal p.d. (first case) $=\boxed{8.0}\text{ V}$
• (d) Power dissipated in $r$ (second case) $=\boxed{1.0}\text{ W}$