Question

An electrical circuit, shown below, consists of a battery, an ammeter, three resistors and two keys.
Consider two cases:
(i) The key $K_1$ is closed and the key $K_2$ is open.
(ii) The key $K_2$ is closed and the key $K_1$ is open.
The ratio of respective currents in these two cases will be

• A. $3:4$
• B. $4:3$
• C. $4:5$
• D. $5:4$

Answer 1

The correct option is D $5:4$

Case $1$: When key $k_1$ is closed and key $K_2$ is open
$\frac{1}{R_{eq}}=\frac{1}{3}+\frac{1}{12}$
$\frac{1}{R_{eq}}=\frac{4+1}{12}$
$R_{eq}=\frac{12}{5}\Omega$
$I_1=\frac{12}{\frac{12}{5}}$
$I_1=5A$

Case $2$: When key $K_2$ is closed and key $k_1$ is open
$\frac{1}{R_{eq}}=\frac{1}{4}+\frac{1}{12}$
$\frac{1}{R_{eq}}=\frac{3+1}{12}$
$R_{eq}=\frac{12}{4}$
$R_{eq}=3\Omega$
$V=I_{2}R$
$I_2=\frac{12}{3}$
$I_2=4A$
$I_1:I_2=5:4$.