Question 583: Current Electricity

Question

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

Accepted Answer

Condition for balanced wheatstone-bridge,
$$\begin{array}{ll} 
& \frac{P}{Q}=\frac{R}{S} \
\Rightarrow & \frac{5}{1}=\frac{50}{10} \
\Rightarrow & \frac{5}{1}=\frac{5}{1}
\end{array}$$
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,
$$V_{P Q}=l_1(5+1)=6 l_1\quad 	ext{.... (i)}$$
and $$V_{R S}=(50+10)\left(1.1-l_1ight)\quad 	ext{... (ii)}$$
Since, for the parallel branches, the voltage drop across them remains same,
$$\begin{aligned}
& 	ext { So, } \
& V_{P Q}=V_{R S} \
& \Rightarrow \quad 6 I_1=60\left(1.1-I_1ight) 	ext { (using Eqs. (i) and (ii)) } \
& \Rightarrow \quad I_1(60+6)=66 \
& \Rightarrow \quad I_1=1 \mathrm{~A}
\end{aligned}$$

Hence, current of 1 A passes through $1 \Omega$ resistance.

Answer

1 A

Answer

0.5 A

Answer

1.1 A

Answer

0.8 A

Current Electricity

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

Choose the correct answer: