Question

A generator develops a maximum voltage of 150 V. A 60 W light bulb is placed across the generator. A maximum current of 0.70 A flows through the bulb.

A) What is the effective rms voltage?

B) What effective rma current flows through the bulb?

C) What is the resistance of the light bulb when it is working?

Answer 1

Part a.

The effective voltage is:

$V_{rms}=\frac{V_{max}}{\sqrt{2}}=\frac{150V}{\sqrt{2}}=106.1V$

Part b

The effective current is:

$I_{rms}=\frac{I_{max}}{\sqrt{2}}=\frac{0.7A}{\sqrt{2}}=0.49A$

Note:

The power developed by the bulb is:$P=V_{rms}*I_{rms}$

Where:

• Voltage $V_{rms}=106.1V$
• Current $I_{rms}=0.49A$

Numerically evaluating: $P=106.1V*0.49A=51.99W\approx 52.0W$

The operating power is below the nominal value of the bulb.

Part c

Resistance is calculated by applying Ohm's Law.

$R=\frac{V_{rms}}{I_{rms}}$

Where:

• Voltage

$V_{rms}=106.1V$

• Electric current

$I_{rms}=0.49A$

Numerically evaluating:

$R=\frac{106.1V}{0.49A}=216.5\Omega$

The resistance of the bulb in operation is

$R=216.5\Omega$