Question

In the circuit shown in Figure 19.77 the emf of the battery is $\mathbf{7}\mathbf{.}\mathbf{4}\mathbf{}\mathbf{V}$. Resistor ${\mathit{R}}_{\mathbf{1}}$has a resistance of $\mathbf{31}\mathbf{}\mathbf{\Omega }$, resistor ${\mathit{R}}_{\mathbf{2}}$ has a resistance of $\mathbf{47}\mathbf{}\mathbf{\Omega }$, and resistor ${\mathit{R}}_3$has a resistance of $\mathbf{52}\mathbf{}\mathbf{\Omega }$ . A steady current flows through the circuit.

(a)What is the equivalent resistance of ${\mathit{R}}_{\mathbf{1}}$and ${\mathit{R}}_{\mathbf{2}}$ ? (b) What is the equivalent resistance of all three resistors? (c) What is the conventional current through${\mathit{R}}_3$

Short answer

0.10 A

Step-by-step solution

Given Data

$\begin{array}{l}{R}_1=31\mathrm{\Omega }\\ R_2=47\mathrm{\Omega }\\ R_3=52\mathrm{\Omega }\\ \mathrm{emf}=7.4\mathrm{V}\end{array}$

Concept

The equivalent resistance is the addition of the resistances, when the batteries are connected in series.

Step 3(c): Determine the conventional current through  R3

Current in the circuit,

$$\begin{gathered} \begin{array}{rcl}i& =& \frac{\mathrm{emf}}{R}\\ & =& \frac{7.4}{70.68}\\ & =& 0.10\mathrm{A}\\ & & \left(\text{Referring to subpart}\left(\text{b}\right)\text{of the SID:875865 - 19 - 56P - b}, \\ \text{take}\mathrm{R}\text{value}\right)\end{array} \end{gathered}$$

Hence, the conventional current through $R_3$ is 0.10 A