Chapter 23, Problem 56A

a.

To determine

Total resistance of load.

Answer to Problem 56A

Total resistance of load is, $R=62\Omega$ .

Explanation of Solution

Given:

Resistances are given;

  $R_1=15\Omega \text{, }{R}_2=47\Omega$

Formula used:

For series combination equivalent resistance is given by,

  $R=R_1+R_2+R_3+\mathrm{.......}{R}_n$

Where, $R$ is equivalent resistance and $R_1,R_2,R_3\mathrm{.......}{R}_n$ are individual resistances.

Calculation:

Now, substituting the value of $R_1,R_2$ and solve.

  $\begin{array}{c}R=15+47\\ =62\Omega \end{array}$

Conclusion:

Therefore, total resistance of load is $62\Omega$ .

b.

To determine

Potential difference across the battery.

Answer to Problem 56A

Potential difference the battery is, $V=6.014\text{ V}$ .

Explanation of Solution

Given:

Resistances are given;

  $R_1=15\Omega \text{, }{R}_2=47\Omega$

Current in the battery is ;

  $\begin{array}{c}i=97\text{ mA}\\ \text{=0}\text{.097 A}\end{array}$

Formula used:

According to ohm’s law.

  $V=iR$

Where, i is current and R is resistance.

Calculation:

Now, substituting the value of i and R and solve.

  $\begin{array}{c}V=0.097\times 62\\ =6.014\text{ V}\end{array}$

Conclusion:

Therefore, potential difference across battery is 6.014 V.