Question.
Compare the power used in the $2 \Omega$ resistor in each of the following circuits:
(i) a $6 \mathrm{~V}$ battery in series with $1 \Omega$ and $2 \Omega$ resistors, and
(ii) a $4 \mathrm{~V}$ battery in parallel with $12 \Omega$ and $2 \Omega$ resistors.
Compare the power used in the $2 \Omega$ resistor in each of the following circuits:
(i) a $6 \mathrm{~V}$ battery in series with $1 \Omega$ and $2 \Omega$ resistors, and
(ii) a $4 \mathrm{~V}$ battery in parallel with $12 \Omega$ and $2 \Omega$ resistors.
solution:
(i) Since $6 \mathrm{~V}$ battery is in series with $1 \Omega 2$ and $2 \Omega \Omega$ resistors, current in the circuit,
$I=\frac{6}{1+2}=\frac{6}{3}=2 \mathrm{~A}$
Power used in $2 \Omega$ resistor, $P_{1}=I^{2} R=(2)^{2} \times 2=8 W$
(ii) Since $4 \mathrm{~V}$ battery is in parallel with $12 \Omega$ and $2 \Omega$ resistors, potential difference across $2 \Omega$ resistor, $V=4 \mathrm{~V}$.
Power used in $2 \Omega$ resistor, $P_{2}=\frac{V^{2}}{R}=\frac{(4)^{2}}{(2)}=8 \mathrm{~W}$
Clearly, $\frac{\mathrm{P}_{1}}{\mathrm{P}_{2}}=\frac{8}{8}=1$
(i) Since $6 \mathrm{~V}$ battery is in series with $1 \Omega 2$ and $2 \Omega \Omega$ resistors, current in the circuit,
$I=\frac{6}{1+2}=\frac{6}{3}=2 \mathrm{~A}$
Power used in $2 \Omega$ resistor, $P_{1}=I^{2} R=(2)^{2} \times 2=8 W$
(ii) Since $4 \mathrm{~V}$ battery is in parallel with $12 \Omega$ and $2 \Omega$ resistors, potential difference across $2 \Omega$ resistor, $V=4 \mathrm{~V}$.
Power used in $2 \Omega$ resistor, $P_{2}=\frac{V^{2}}{R}=\frac{(4)^{2}}{(2)}=8 \mathrm{~W}$
Clearly, $\frac{\mathrm{P}_{1}}{\mathrm{P}_{2}}=\frac{8}{8}=1$
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