(a) State Joules law of heating and write its mathematical expression.

(b) Two resistors of resistances 2 Ω and 4 Ω are connected in

1. series
2. parallel

with a battery of given potential difference. Compute the ratio of total quantity of heat produced in the combination in the two cases if the total voltage and time are kept the same for both.

(a) Joule's Law of Heating states that the amount of heat produced in a conductor is directly proportional to the square of the electric current passing through it, the resistance of the conductor and the time for which the current flows.

Mathematically, it can be expressed as

$\text H = \text I^2 \text {Rt}$

• $\text H$ is the heat produced (in joules),
• $\text I$ is the electric current (in amperes),
• $\text R$ is the resistance of the conductor (in ohms),
• $\text t$ is the time for which the current flows (in seconds).

(b) Given,

• $\text R_1$ = 2 Ω
• $\text R_2$ = 4 Ω
• $\text t$1 = \text t2 = \text t(let)
• $\text V$1 = \text V2 = \text V(let)

1. Series connection

Net resistance of the circuit is given by,

$\text R$\text S = \text R1 + \text R_2 \\[1em] = 2 + 4 \\[1em] = 6\ \text Ω

Let current through the combination is $\text I_\text S$.

Then,

$\text I$\text S = \dfrac{\text V1}{\text R_\text S} \\[1em] = \dfrac{\text V}{6}

So, heat produced by this combination is given by,

$\text H$\text S = \text I\text S^2 \text R\text S \text t1 \\[1em] = \left(\dfrac{\text V}{6}\right)^2 \times 6 \times \text t \\[1em] = \dfrac{\text V^2}{36} \times 6 \times \text t \\[1em] = \dfrac{\text V^2}{6} \times \text t \\[1em] \Rightarrow \text H_\text S = \dfrac{\text V^2 \text t}{6}

2. Parallel connection

Net resistance of the circuit is given by,

$\dfrac {1}{\text R$\text P} = \dfrac {1}{\text R1} + \dfrac {1}{\text R2} \\[1em] = \dfrac{1}{2} + \dfrac{1}{4} \\[1em] = \dfrac{2 + 1}{4} \\[1em] = \dfrac{3}{4} \\[1em] \Rightarrow \text R\text P = \dfrac{4}{3}\ \text Ω

Let current through the combination is $\text I_\text P$.

Then,

$\text I$\text P = \dfrac{\text V2}{\text R_\text P} \\[1em] = \dfrac{\text V}{\dfrac{4}{3}} \\[1em] = \dfrac{3 \times \text V}{4} \\[1em] = \dfrac{3\text V}{4}

So, heat produced by this combination is given by,

$\text H$\text P = \text I\text P^2 \text R\text P \text t2 \\[1em] = \left(\dfrac{3\text V}{4}\right)^2 \times \dfrac{4}{3} \times \text t \\[1em] = \dfrac{9\text V^2}{16} \times \dfrac{4}{3} \times \text t \\[1em] = \dfrac{3\text V^2}{4} \times \text t \\[1em] \Rightarrow \text H_\text P = \dfrac{3\text V^2 \text t}{4}

Now,

$\dfrac{\text H$\text S}{\text H\text P} = \dfrac {\dfrac{\text V^2 \text t}{6}}{\dfrac{3\text V^2 \text t}{4}} \\[1em] = \dfrac{4\text V^2 \text t}{18\text V^2 \text t} \\[1em] = \dfrac {4}{18} \\[1em] = \dfrac{2}{9}

Hence, the ratio of heat energy produced by series connection to the parallel connection is 2 : 9.