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

$V_{R1}=2.71\mathrm{V}$

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(a): Determine the absolute value of the potential difference across R1

First we have to find the current,

$\begin{array}{c}l=\frac{emf}{R_1+R_2}\\ =\frac{7.9}{23+44}\\ =0.1179\mathrm{\Omega }\end{array}$

The potential difference,

$\begin{array}{c}{V}_{R1}=I{R}_1\\ =0.1179\times 23\\ =2.71\mathrm{V}\end{array}$

Hence, the absolute value of the potential difference across resistor 1 is 2.71 V