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 localid="1662203602529" ${\mathit{R}}_3$has a resistance of localid="1662203617219" $\mathbf{52}\mathbf{}\mathbf{\Omega }$. 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

$18.68\mathrm{\Omega }$

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(a): Determine the equivalent resistance of  R1 and R2

The resistors $R_1$ and $R_2$are connected in parallel, then the equivalent resistance,

$\begin{array}{rcl}{R}_{eq}& =& \frac{1}{R_1}+\frac{1}{R_2}\\ & =& \frac{1}{31}+\frac{1}{47}\\ & =& 0.05\mathrm{\Omega }\\ {\mathrm{R}}_4& =& 1/0.05\\ & =& 18.68\mathrm{\Omega }\end{array}$

Hence, the equivalent resistance is $18.68\mathrm{\Omega }$$18.68\mathrm{\Omega }$