Question

When glowing, a thin-filament bulb has a resistance of about $\mathbf{30}\mathbf{}\mathbf{\Omega }$and a thick filament bulb has a resistance of about $\mathbf{10}\mathbf{}\mathbf{\Omega }$. If they are in parallel, what is their equivalent resistance? How much current goes through two $\mathbf{1}\mathbf{.}\mathbf{5}\mathbf{}\mathbf{V}$flashlight batteries in series if a thin-filament bulb and a thick filament bulb are connected in parallel to batteries?

Short answer

The equivalent resistance of the parallel combination of the resistors is$7.5\mathrm{\Omega }$ and the current flowing through the flashlight batteries is $0.4\mathrm{A}$.

Step-by-step solution

Given Data

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

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

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

Concept

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" $\frac{\mathbf{1}}{{\mathbf{R}}_{\mathbf{p}}}\mathbf{=}\frac{\mathbf{1}}{{\mathbf{R}}_{\mathbf{1}}}\mathbf{+}\frac{\mathbf{1}}{{\mathbf{R}}_{\mathbf{2}}}$

Calculations

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

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

Further solving, we get-

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

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

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" $$\begin{gathered} I=\frac{V}{R_p} \\ =\frac{2\times 1.5\mathrm{V}}{7.5\mathrm{\Omega }} \\ =0.4\mathrm{A} \end{gathered}$$

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