91Ó°ÊÓ

Resistance of thin filament bulb:$30\mathrm{Î\copyright }$

Resistance of thick filament bulb:$10\mathrm{Î\copyright }$

The voltage across terminals of a battery:$1.5\mathrm{V}$

When two resistances ${\mathit{R}}_{\mathbf{1}}\mathbf{}\mathbf{and}\mathbf{}{\mathit{R}}_{\mathbf{2}}$ are connected in parallel, the reciprocal of their equivalent resistance${\mathit{R}}_{\mathbf{p}}$ is equal to the sum of the reciprocal of the two resistances.

role="math" localid="1662123335573" $\frac{\mathbf{1}}{{\mathbf{R}}_{\mathbf{p}}}\mathbf{=}\frac{\mathbf{1}}{{\mathbf{R}}_{\mathbf{1}}}\mathbf{+}\frac{\mathbf{1}}{{\mathbf{R}}_{\mathbf{2}}}$

The equivalent resistance of the two bulbs can be calculated using equation.

$\frac{1}{R_p}=\frac{1}{R_1}+\frac{1}{R_2}$

For $R_1=30\mathrm{Î\copyright }$and$R_2=10\mathrm{Î\copyright }$, the equation becomes-

role="math" localid="1662123653974" $$\begin{gathered} \frac{1}{R_p}=\frac{1}{30\mathrm{Î\copyright }}+\frac{1}{10\mathrm{Î\copyright }} \\ =\frac{1+3}{30\mathrm{Î\copyright }} \\ =\frac{4}{30\mathrm{Î\copyright }} \end{gathered}$$

Further solving, we get-

role="math" localid="1662123810780" $$\begin{gathered} R_p=\frac{30}{4}\mathrm{Î\copyright } \\ =7.5\mathrm{Î\copyright } \end{gathered}$$

The equivalent resistance of the parallel combination of two bulbs is: $7.5\mathrm{Î\copyright }$

When two batteries are connected in series combination, parallel to the combination of resistors, the current flowing in the circuit will be role="math" localid="1662123905440" $I$.

Using Ohm’s Law, the current in the circuit is given as-

role="math" localid="1662124013334" $$\begin{gathered} I=\frac{V}{R_p} \\ =\frac{2×1.5\mathrm{V}}{7.5\mathrm{Î\copyright }} \\ =0.4\mathrm{A} \end{gathered}$$

The current flowing through the batteries is role="math" localid="1662124063254" $0.4\mathrm{A}$.