In the circuit given below, if the bulb is to glow with maximum intensity, the value of 'R' is (neglect internal resistance of the cell)
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For maximum intensity, the bulb should operate at its rated voltage and power. Calculate the bulb's resistance. Use Kirchhoff's voltage law and current division rules to find the value of \( R \) that results in the rated voltage across the bulb.
The bulb is rated at 1. 5 V and 0. 45 W. To glow with maximum intensity, the voltage across the bulb should be 1. 5 V and the power dissipated should be 0. 45 W.
We can find the resistance of the bulb \( R_{bulb} \) using the power rating:
$$ P = \frac{V^2}{R} \implies R_{bulb} = \frac{V^2}{P} = \frac{(1. 5)^2}{0. 45} = \frac{2. 25}{0. 45} = 5 \, \Omega $$
The circuit consists of a 3 \( \Omega \) resistor in series with a parallel combination of resistor \( R \) and the bulb with resistance \( R_{bulb} = 5 \, \Omega \). The total voltage of the cell is 6 V.
For the voltage across the parallel combination (and hence across the bulb) to be 1. 5 V, the voltage drop across the 3 \( \Omega \) resistor must be \( 6 - 1. 5 = 4. 5 \) V.
Let the current through the 3 \( \Omega \) resistor be \( I \). Then, \( V_{3\Omega} = I \times 3 = 4. 5 \) V, which gives \( I = \frac{4. 5}{3} = 1. 5 \) A.
This current \( I \) is the total current flowing from the cell. It divides into two branches in the parallel combination: current through \( R \) (\( I_R \)) and current through the bulb (\( I_{bulb} \)).
The current through the bulb can be found using its power rating and voltage:
$$ P = V I \implies I_{bulb} = \frac{P}{V} = \frac{0. 45}{1. 5} = 0. 3 \, \text{A} $$
Now, the current through \( R \) is \( I_R = I - I_{bulb} = 1. 5 - 0. 3 = 1. 2 \, \text{A} \).
Since the voltage across \( R \) is the same as the voltage across the bulb (parallel combination), which is 1. 5 V, we can find the value of \( R \) using Ohm's law:
$$ V_R = I_R R \implies R = \frac{V_R}{I_R} = \frac{1. 5}{1. 2} = \frac{15}{12} = \frac{5}{4} = 1. 25 \, \Omega $$
The value of \( R \) required for the bulb to glow with maximum intensity is 1. 25 \( \Omega \).
