Question

In the circuit shown in Figure 19.75, the emf of the battery is $\mathbf{7}\mathbf{.}\mathbf{9}\mathbf{}\mathbf{V}$. Resistor ${\mathit{R}}_{\mathbf{1}}$ has a resistance of $\mathbf{23}\mathbf{}\mathbf{\Omega }$, and resistor ${\mathit{R}}_{\mathbf{2}}$has a resistance of$\mathbf{44}\mathbf{}\mathbf{\Omega }$. A steady current flows through the circuit. (a) What is the absolute value of the potential difference across ${\mathit{R}}_{\mathbf{1}}$? (b) What is the conventional current through ${\mathit{R}}_{\mathbf{2}}$?

Short answer

$l=0.1179\mathrm{\Omega }$

Step-by-step solution

Given Data

$\begin{array}{l}\mathrm{emf}=7.9\mathrm{V}\\ {\mathrm{R}}_1=23\mathrm{\Omega }\\ {\mathrm{R}}_2=44\mathrm{\Omega }\end{array}$

Concept

If the current flows through the source of the negative terminal, then it is known as the conventional current.

Step 3(b): Determine the conventional current through R2

As per Kirchhoff’s current Law,

$\begin{array}{c}-7.9\mathrm{V}+I{R}_1+I{R}_2=0\\ -7.9\mathrm{V}=-I\left(R_1+R_2\right)\\ -7.9\mathrm{V}=-I\left(23+44\right)\\ -7.9\mathrm{V}=-67I\\ I=0.1179\mathrm{\Omega }\end{array}$

Hence, the conventional current through is $0.1179\mathrm{\Omega }$