Question

In the circuit shown, n identical resistors R are connected in parallel (n>1) and the combination is connected in series to another resistor $R_0$. In the adjoining circuit n resistors of resistance R are all connected in series along with $R_0$ .
The batteries in both circuits is same and the net power dissipated in the n resistors in both the circuits is the same. The ratio $R_0/R$ is

• A. $1$
• B. $n$
• C. $n^2$
• D. $1/n$

Answer 1

The correct option is A $1$

In the parallel circuit $\frac{1}{R_{eq1}}=\frac{n}{R}$

$R_{eq1}=\frac{R}{n}+R_o$

In series $R_{eq2}=R_o+nR$

The batteries in both the circuits is the same'

$V_1$=$V_2$

$I_1{R}_{eq1}$ $=$ $I_2{R}_{eq2}$.....(1)

Now, the power dissipated across n resistors is the same in the two circuits:

$P_1$ $=$ $P_2$

Therefore, $I_1^{2}R/n$ $=$ $I_2^{2}nR$

$(I_1/I_2{)}^2$ $=$ $n^2$

or $I_1/I_2$ $=$ $n$

using the above relation in equation 1, we get,

$n$ $=$ {$R_o+nR$}$/${$R/n+R_o$}

solving the above equation we get, $R_o/R$ $=$ 1