Chapter 23.1, Problem 3PP

To determine

The resistance of each bulb in a given series circuit.

Answer to Problem 3PP

The resistance of each bulb in the circuit is $195\Omega$.

Explanation of Solution

Given:

A series of ten identical bulbs with equal resistance values connected to a 117-V outlet and current through the circuit is 0.06 A.

Formula used:

Given the current flowing through a circuit and supply voltage, the total resistance can be calculated using the Ohm’s law as,

  $\begin{array}{l}V=IR\\ R=\frac{V}{I}\end{array}$ …… (1)

Also, the equivalent resistance of the series resistors is the sum of individual resistance values.

  $R_{eq}=R_A+R_B+\mathrm{...}$ ……. (2)

Calculation:

For the series circuit with a voltage of 110 V and current 0.06 A, total resistance of the light circuit can be calculated by using the relation in (1) as

  $\begin{array}{c}R=\frac{117}{0.06}\\ =1950\Omega \end{array}$

Now, since the circuit consists of ten identical bulbs in series the above resistance value, the resistance of each bulb can be calculated using (2) as,

  $\begin{array}{c}{R}_{eq}=10R\\ R=\frac{R_{eq}}{10}\\ =\frac{1950}{10}\\ =195\Omega \end{array}$

Conclusion:

Therefore, the resistance of each bulb is $195\Omega$ .