Question

Calculate the value of the resistance $R$ in the circuit shown in the figure so that the current in the circuit is $0.2A$. What would be the potential difference between points $B$ and $E$?

Answer 1

$\text{Step 1: Simplify the given circuit by combining Series and Parallel Resistances}$

First, combining series Resistances $5\Omega$ and $10\Omega$ we get $15\Omega$ resistance, as shown in Figure.

The remaining three resistances in the reduced circuit are in parallel, So

$\frac{1}{R_{BE}}=\frac{1}{30}+\frac{1}{10}+\frac{1}{15}=\frac{1}{5}$

$\Rightarrow R_{BE}=5\Omega$

$\text{Step 2: Apply Kirchhoff's Law}$

The Reduced circuit is as follows:

Now, Applying KVL in clockwise direction in the loop, increase in voltage is written as positive and decrease as negative, So:

$8V-0.2A\times 5\Omega -0.2A\times R-3V-15\Omega \times 0.2A=0$

$\Rightarrow R=\frac{8-0.2\times 5-15\times 0.2-3}{0.2}$ $=5\Omega$

$\text{Step 3: Solving Equations to find the required value}$

To find Voltage across Points BE, refer figure below:

$\therefore V_{BE}=I{R}_{BE}=5\times 0.2=1V$

Hence, the Potential difference between points B and E is $1V$.