D.C. Circuits: Question 3

Worked solution

Part (a): Kirchhoff’s second law and the current

Kirchhoff’s second law states that, around any complete loop, the sum of the e.m.f.s equals the sum of the potential drops (a consequence of conservation of energy). Going around this single-loop circuit: $E = IR + Ir = I(R+r)$

Rearranging for $I$ and substituting $E=9.0\text{ V}$, $R=4.0\ \Omega$, $r=0.50\ \Omega$: $I = \frac{E}{R+r} = \frac{9.0}{4.0+0.50} = \frac{9.0}{4.5} = 2.0\text{ A}$

Check: $4.5\times2.0=9.0\text{ V}$, which matches $E$. Confirmed.

Part (b): Terminal potential difference

The terminal p.d. is the potential difference actually available across the battery’s terminals, the p.d. across the external resistor $R$: $V_{\text{terminal}} = IR = 2.0\times4.0 = 8.0\text{ V}$

Check by an independent route: the terminal p.d. also equals $E-Ir=9.0-(2.0\times0.50)=9.0-1.0=8.0\text{ V}$. Both routes agree.

Part (c): Potential difference across the internal resistance (lost volts)

The “lost volts” is the potential difference across the internal resistance itself: $V_{\text{lost}} = Ir = 2.0\times0.50 = 1.0\text{ V}$

Check: the terminal p.d. and the lost volts should add up to the e.m.f.: $8.0+1.0=9.0\text{ V}=E$. Confirmed.

Part (d): Power dissipated in $r$ and efficiency

The power dissipated in the internal resistance is: $P_r = I^2r = (2.0)^2\times0.50 = 4.0\times0.50 = 2.0\text{ W}$

The total power supplied by the battery is: $P_{\text{total}} = EI = 9.0\times2.0 = 18.0\text{ W}$

Check using the external resistor: the useful power delivered to $R$ is $P_R=I^2R=(2.0)^2\times4.0=16.0\text{ W}$, and $P_R+P_r=16.0+2.0=18.0\text{ W}=P_{\text{total}}$. Confirmed. Energy is conserved between the two resistances.

The efficiency of the energy transfer to the external resistor is the ratio of useful power output to total power supplied: $\text{efficiency} = \frac{P_R}{P_{\text{total}}} = \frac{16.0}{18.0} = 0.8889$

Check as a fraction: $\frac{16.0}{18.0}=\frac{8}{9}=0.8889$, the same value. So the efficiency is $\boxed{88.9}\%$ (3 s.f.).

Final answers

• (a) $E=I(R+r)$, giving $I=\boxed{2.0}\text{ A}$
• (b) Terminal p.d. $=\boxed{8.0}\text{ V}$
• (c) Lost volts $=\boxed{1.0}\text{ V}$
• (d) Power dissipated in $r=\boxed{2.0}\text{ W}$; efficiency $=\boxed{88.9}\%$