Chapter 23, Problem 10STP

To determine

To Calculate: The current and power in the circuit.

Answer to Problem 10STP

  $\text{0}\text{.2 A; 1}\text{.6 W}$

Explanation of Solution

Given:

Value of resistance, $R_1=4.0\Omega$

Value of resistance, $R_2=8.0\Omega$

Value of resistance, $R_3=13.0\Omega$

Value of resistance, $R_4=15.0\Omega$

Voltage of battery, $V=8.0\text{V}$

Formula used:

Equivalent resistance for series combination can be obtained by:

  $R_{eq}=R_1+R_2+R_3+\mathrm{...}$

Ohm’s law:

  $V=IR$

Here, I is the current, V is the potential difference and R is the resistance.

Power in the circuit can be obtained by:

  $P=IV$

Calculations:

  $\begin{array}{c}{R}_{eq}=4.0+8.0+13.0+15.0\\ =40\Omega \end{array}$

Using Ohm’s law, current in the circuit is:

  $\begin{array}{c}I=\frac{V}{R}\\ =\frac{8.0}{40}\\ =0.2\text{A}\end{array}$

Power in the circuit is:

  $\begin{array}{c}P=\left(0.2\right)\left(8.0\right)\\ =1.6\text{W}\end{array}$

Conclusion:

Thus, the current in the circuit is 0.2 A and power is 1.6 W.