Question

In a circuit, 0.5 A is flowing through the bulb. The voltage across the bulb is 4.0 V. What is the bulb
resistance?

Answer 1

Energy (power×time) is measured in Joules and by including time (t) in the power formulae, the energy dissipated by a component or circuit can be calculated.

Energy dissipated = Pt or VI×tasP=VI.

In the question, it is given that the current I is 0.5 A and voltage V is 4 V. Thus the power dissipated is calculated as P=VI=4V×0.5A=2W.

Substituting V=IR (Ohm's law) in the above formula, we get, P=I

2

×R.

Therefore, the total resistance in the circuit is calculated using the relation R=

I

2

P

=

0.5

2

2

=8ohms.

The internal resistance of the cell is 2.5 ohms, hence, the resistance of the bulb is 8−2.5=5.5 ohms.

The power dissipated by the bulb in 10 minutes is calculated from the formula P=I

2

×R×t, where I=0.5A,R=5.5 ohms and t=10 minutes, that is 600 seconds.

Hence, P=0.5

2

×5.5×600=825J.

The power dissipated by the bulb in 10 minutes is 825 Joules.

Answer 2

Given :

In a circuit,

Current flowing through the bulb = 0.5 A

Potential difference = 4 volts

To find :

The resistance of the bulb

Solution :

Using ohms law that is,

» At constant temperature the current flowing through a conductor is directly proportional to the potential difference across its ends.

Formula : V = RI

where,

• V denotes potential difference
• R denotes resistance
• I denotes current

substituting all the given values in the formula,

$\dashrightarrow \sf V = RI$

$\dashrightarrow \sf 4 = R(0.5)$

$\dashrightarrow \sf R = \dfrac{4}{0.5}$

$\dashrightarrow \sf R = 8 \: ohms$

Thus, the resistance of the bulb is 8 ohms.