Question

a circuit consistsof 1 ohm wire connected in series with parallel arrangement of 6 ohm and 3 ohm.calculate the total resistance of circuit

Answer 1

CORRECT QUESTION :-

➠ A circuit consists of 1 ohm wire connected in series with parallel arrangement of 6 ohm and 3 ohm. Calculate the total resistance of the circuit .

TO FIND :-

➠ Total resistance of the circuit .

SOLUTION :-

✮ Formula to find the equivalent resistance when the resistors are arranged in parallel combination .

$\dagger \: \boxed{ \red{ \bf \: \dfrac{1}{ R_{eq} } = \dfrac{1}{R_{1} } + \dfrac{1}{ R_{2} } + ..... + \dfrac{1}{ R_{n} } }}$

✮ Formula to find the equivalent resistance when the resistors are arranged in series combination .

$\dagger \boxed{ \red{ \bf \:R_{eq } = R_{1} + R_{2} + R_{3} + .... + R_{n}}}$

➠ Now let's calculate the equivalent resistance in parallel arrangement .

$\to \: \boxed{ \rm \: \: \dfrac{1}{ R_{eq} } = \dfrac{1}{R_{1} } + \dfrac{1}{ R_{2} } + ..... + \dfrac{1}{ R_{n} }} \\ \\ \to \: \rm \dfrac{1}{ R_{eq} } = \dfrac{1}{6 \Omega} + \dfrac{1}{3\Omega} \\ \\ \to \rm \: \dfrac{1}{ R_{eq} } = \dfrac{3}{18 \Omega} + \dfrac{6}{18 \Omega} \\ \\ \to \rm \: \dfrac{1}{ R_{eq} } = \dfrac{3 + 6}{18 \Omega} \\ \\ \to \rm \: \dfrac{1}{ R_{eq} } = \dfrac{ \cancel{18}\Omega}{ \cancel{9} } \\ \\ \to\boxed{ \red{ \rm \: R_{eq} = 2 \Omega}}$

➠ Hence the equivalent resistance between the parallel arrangement is 2 ohm.

➠ Now let's calculate the net resistance of the circuit.

$\to \rm \: R_{net} = 2 \Omega + 1 \Omega \\ \\ \to \boxed{ \red{ \rm \: R_{net} = 3\Omega}}$

Hence the net resistance of the circuit is 3 ohm.