Question

In the electric circuit shown in the figure, the potential difference across 3 ohm resistance will be:
(a) 12V (c) 2.4V
(b)24V (d) 36V
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Answer 1

hey mate, here is your answer.

Answer 2

Answer:

The potential difference across 3Ω resistance = 24V

Explanation:

Let $R_1$ = 3Ω

$R_2$ = 6Ω

$R_3$ = 4Ω

$R_4$ = 12Ω

Let $R_p$ is the combined resistance of  $R_1$ and $R_2$

Since the resistances 3Ω and 6Ω are connected in parallel, then the total resistance

$R_p$ = $\frac{R_1XR_2}{R_1+R_2}$

= $\frac{3X6}{3+6}$

=2Ω

$R_p$ =2Ω

Let $R_q$ is the combined resistance of $R_3$ and $R_p$

Since the resistances $R_3$ and $R_p$are connected in series,

then the total resistance

$R_q$ =  $R_3$ + $R_p$

= 4Ω+2Ω

= 6Ω

$R_q$ = 6Ω

Let R is the combined resistance of  $R_q$ and $R_4$

Since the resistances  $R_q$ and $R_4$ are connected in parallel, then the combined resistance

R = $\frac{R_qXR_4}{R_q+R_4}$

= $\frac{6X12}{6+12}$

=4Ω

R = 4Ω

∴ The total resistance across the circuit = 4Ω

Given, that the current in circuit = 18A

We know,

The Potential difference V = IR, where V is the potential difference, I is the current and R is the resistance

V = 18 × 4 = 72V

Since the potential difference across the parallel connection is the same, the current across  $R_q$

$I_q$ = $\frac{V}{R_q}$

= $\frac{72}{6}$

= 12A

The current across  $R_q$ =  12A

Then potential difference across $R_P$

V =  I ×  $R_P$

=12 × 2

= 24V

Since the potential difference across the parallel connection remains same.

The potential difference across 3Ω = The potential difference across  $R_P$

= 24V

∴The potential difference across 3Ω resistance = 24V

#SPJ3