Consider the following electric circuit :

Calculate the values of the following :

(a) The total resistance of the circuit

(b) The total current drawn from the source

(c) Potential difference across the parallel combination of 10 Ω and 15 Ω resistors

(a) Given,

• Battery voltage ($\text V$) = 15 V

From the figure,

10 Ω and 15 Ω resistors are in parallel then their total resistance is given by,

$\dfrac{1}{\text R$1} = \dfrac{1}{10} + \dfrac{1}{15} \\[1em] = \dfrac{3 + 2}{30} \\[1em] = \dfrac{5}{30} \\[1em] = \dfrac{1}{6} \\[1em] \Rightarrow \text R1 = 6\ \text Ω

Again, 40 Ω and 60 Ω are also in parallel then their total resistances given by,

$\dfrac{1}{\text R$2} = \dfrac{1}{40} + \dfrac{1}{60} \\[1em] = \dfrac{6 + 4}{240} \\[1em] = \dfrac{10}{240} \\[1em] = \dfrac{1}{24} \\[1em] \Rightarrow \text R2 = 24\ \text Ω

Now, $\text R$1 and $\text R2$ are in series then total resistance of the circuit is given by,

$\text R$\text S = \text R1 + \text R2 \\[1em] = 6 + 24 \\[1em] \Rightarrow \text R\text S = 30\ \text Ω

Hence, the total resistance of the circuit is 30 Ω.

(b) Current drawn from the battery is given by,

$\text I = \dfrac{\text V}{\text R_\text S} \\[1em] = \dfrac{15}{30} \\[1em] = \dfrac{1}{2} \\[1em] \Rightarrow \text I = 0.5\ \text A$

Hence, the total current drawn from the source is 0.5 A.

(c) Potential difference across the parallel combination of 10 Ω and 15 Ω resistors is given by,

$\text V$1 = \text {IR}1 \\[1em] = 0.5 \times 6 \\[1em] \Rightarrow \text V_1 = 3\ \text V

Hence, the potential difference across the parallel combination of 10 Ω and 15 Ω resistors is 3 V.