Question

The current in $1\Omega$ resistor in the following circuit is

• A. $1A$
• B. $0.5A$
• C. $1.1A$
• D. $0.8A$

Answer 1

Answer: (A)

Condition for balanced wheatstone-bridge,
$\frac{P}{Q}=\frac{R}{S}$
$\Rightarrow \frac{5}{1}=\frac{50}{10}$
$\Rightarrow \frac{5}{1}=\frac{5}{1}$
So, it is a balanced wheatstone-bridge and hence no current will flow through the arm $G$. The, circuit can be redrawn as given below,

As, we know that in a series branch, same current flows.
Hence, from the KVL loop rule,
$V_{PQ}=I_1(5+1)=6l_1$...(i)
and $V_{RS}=(50+10)\left(1.1-l_1\right)$(ii)
Since, for the parallel branches, the voltage drop across them remains same, So, $V_{PQ}=V_{RS}$
$\Rightarrow 6l_1=60\left(1.1-l_1\right)$ (using Eqs. (i) and(ii))
$\Rightarrow l_1(60+6)=66$
$\Rightarrow l_1=1A$
Hence, current of 1 A passes through $1\Omega$ resistance.