The circuit diagram given below shows the combination of three resistors $ R_{1}, R_{2} $ and $ R_{3} $ :
Find: (i) total resistance of the circuit.
(ii) total current flowing in the circuit.
(iii) the potential difference across $ R_{1} $."

Given, 

Potential difference, V = 12V

Resistors, R₁ = 4Ω, R₂ = 3Ω, R₃ = 6Ω

(i) To find out the total resistance of the circuit we need to add R₁ + R₂ + R₃.

As shown in the figure, the resistors R₂ and R₃ are connected in parallel. So, first we add R₂ and R₃, and their total resistance is given as:

$\frac{1}{R}=\frac{1}{{R}_{2}}+\frac{1}{{R}_{3}}$

$\frac{1}{R}=\frac{1}{3}+\frac{1}{6}$

$\frac{1}{R}=\frac{2+1}{6}$

$\frac{1}{R}=\frac{3}{6}$

$\frac{1}{R}=\frac{1}{2}$

$R=2\Omega $

Now, this resistance is in the series with the resistor, R₁.

Hence, the total resistance in the circuit is given as:

${R}_{T}={R}_{1}+({R}_{2}+{R}_{3})$

${R}_{T}=4+2$     $\left[\because {R}_{1}=4\Omega \ and\ ({R}_{2}+{R}_{3})=2\Omega \right]$

${R}_{T}=6\Omega $ 

Thus, the total resistance of the circuit is 2 Ampere.

(ii) Total current flowing through the circuits can be calculated as follows:

$I=\frac{V}{R}$,  where I = current, V = potential difference, and R = resistance

Substituting the values in the equation, we get-

$I=\frac{12}{6}$

$I=2A $ 

Thus, the total current flowing through the circuit is 2 Ampere.

(ii) The potential difference (V) across R₁ is given as:

$V=R\times I$

Substituting the values we get:

$V=4\Omega \times 2A$

$V=8V$

Thus, the potential difference (V) across R₁ is 8 Volt.

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Updated on: 10-Oct-2022

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