Question

Two resistors of resistance
2 ohms and 4 ohms are
connected in parallel to a 6 volt
battery. Find the heat dissipated
in the 4 ohm resistor in 5 seconds

Answer 1

Given :

• Resistance $(R_{1})$ = 2 Ω

• Resistance $(R_{2})$ = 4 Ω

• Potential Difference = 6 V

• Time = 5 s

To Find :

The heat energy produced in the 4 Ω resistor.

Solution :

First we have to find the equivalent resistance of the circuit , so that we can find the current in the circuit.

Formula for Equivalent resistance in a Parallel Circuit.

$\bf{\dfrac{1}{R_{eq}} = \dfrac{1}{R_{1}} + \dfrac{1}{R_{2}} + \dfrac{1}{R_{3}} + ... + \dfrac{1}{R_{n}}}$

Where :

• $R_{eq}$ = Equivalent resistance

• $R$ = Resistance

Now , using the above formula and substituting the values in it , we get :

$:\implies \bf{\dfrac{1}{R_{eq}} = \dfrac{1}{R_{1}} + \dfrac{1}{R_{2}}}$

$:\implies \bf{\dfrac{1}{R_{eq}} = \dfrac{1}{2} + \dfrac{1}{4}}$

$:\implies \bf{\dfrac{1}{R_{eq}} = \dfrac{2 + 1}{4}}$

$:\implies \bf{\dfrac{1}{R_{eq}} = \dfrac{3}{4}}$

$:\implies \bf{R_{eq} = \dfrac{4}{3}}$

$:\implies \bf{R_{eq} = 1.33\:\Omega}$

Hence, the equivalent resistance in the circuit is 1.33 Ω.

Now , using the ohm's law and Substituting the values in it , we get :

$:\implies \bf{V = IR}$

$:\implies \bf{6 = l \times 1.33}$

$:\implies \bf{\dfrac{6}{1.33} = l}$

$:\implies \bf{\dfrac{6}{1.33} = l}$

$:\implies \bf{4.5\:A = l}$

$:\implies \bf{l = 4.5\:A}$

Hence, the current in the circuit is 4.5 A.

To Find the heat produced :

Using the formula for heat produced and Substituting the values in it , we get :

$:\implies \bf{H = I^{2}RT}$

$:\implies \bf{H = 4.5^{2} \times 1.33 \times 5}$

$:\implies \bf{H = 20.25 \times 1.33 \times 5}$

$:\implies \bf{H = 134.66\:J}$

Hence, the heat produced is 134.66 J.