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${\mathit{R}}_{\mathbf{3}}$

Short answer

$70.68\mathrm{\Omega }$

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(b): Determine the equivalent resistance of all three resistors

The equivalent resistance of all the three resistors,

$\begin{array}{rcl}R& =& R_3+R_4\\ & =& 52+18.68\\ & =& 70.68\mathrm{\Omega }\\ & & \left(\text{Referring to subpa}\right)\text{rt}\left(\text{a}\right)\text{of the SID:875865 - 19 - 56P - a},\\ & & \text{take}{\mathrm{R}}_4\text{value}\end{array}$

Hence, the equivalent resistance of all three resistors is $70.68\mathrm{\Omega }$