### Current Electricity Question 1

#### Question 1 - 24 January - Shift 1

As shown in the figure, a network of resistors is connected to a battery of $24 V$ with an internal resistance of $3 \Omega$. The currents through the resistors $R_4$ and $R_5$ are $I_4$ and $I_5$ respectively. The values of $I_4$ and $I_5$ are :

$(24 V, 3 \Omega)$

(1) $I_4=\frac{8}{5} A$ and $I_5=\frac{2}{5} A$

(2) $I_4=\frac{24}{5} A$ and $I_5=\frac{6}{5} A$

(3) $I_4=\frac{6}{5} A$ and $I_5=\frac{24}{5} A$

(4) $I_4=\frac{2}{5} A$ and $I_5=\frac{8}{5} A$

## Show Answer

#### Answer: (4)

#### Solution:

#### Formula: Combination of Resistances

Equivalent resistance of circuit

$ \begin{aligned} R _{eq} & =3+1+2+4+2 \\ & =12 \Omega \end{aligned} $

Current through battery $i=\frac{24}{12}=2 A$

$I_4=\frac{R_5}{R_4+R_5} \times 2=\frac{5}{20+5} \times 2=\frac{2}{5} A$

$I_5=2-\frac{2}{5}=\frac{8}{5} A$