Question

In the circuit given below , calculate :
(a) total effective resistance
(b) current through 6 Ω resistor
(c) potential difference across 4Ω resistor

Answer 1

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Answer 2

Answer: The total effective resistance is 3 ohm, the current through 6 ohm is 1 A and the potential difference across 4 ohm is 4 volt.

Explanation:

Given that,

Resistance $R_{1} = 4\Omega$

Resistance $R_{2} = 2\Omega$

Resistance $R_{3} = 6\Omega$

According to the figure,

$R_{1}$ and $R_{2}$ are connected in series then,

The resistance R is

$R = R_{1}+R_{2}$

$R = 4\Omega+2\Omega$

$R = 6\Omega$

$R$ and $R_{3}$ are connected in parallel

(a). The effective resistance is

$\dfrac{1}{R_{eff}}=\dfrac{1}{R}+\dfrac{1}{R_{3}}$

$\dfrac{1}{R_{eff}}=\dfrac{1}{3}$

$R_{eff} = 3 \Omega$

So, the total effective resistance is 3 ohm.

(b). The current through 6 ohm resistor is

Using ohm's law

$V = i\times R$

$i = \dfrac{V}{R}$

$i = \dfrac{6 V}{6\Omega}$

$i = 1 A$

(c). The potential difference across 4 ohm resistor

$V = i\times R$

$V = 1 A\times 4\Omega$

$V = 4 volt$

The potential difference is 4 volt.

Hence, The total effective resistance is 3 ohm, the current through 6 ohm is 1 A and the potential difference across 4 ohm is 4 volt.