In the circuit given below,if the bulb is to | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) The bulb is rated at $V_b = 1.5 	ext{ V}$ and $P = 0.45 	ext{ W}$.The resistance of the bulb is $R_b = \frac{V_b^2}{P} = \frac{(1.5)^2}{0.45} =

Vedclass · Accepted Answer

$1.25$

Vedclass · Answer

(A) The bulb is rated at $V_b = 1.5 	ext{ V}$ and $P = 0.45 	ext{ W}$.The resistance of the bulb is $R_b = \frac{V_b^2}{P} = \frac{(1.5)^2}{0.45} = \frac{2.25}{0.45} = 5 \Omega$.For the bulb to glow with maximum intensity,it must operate at its rated voltage of $1.5 	ext{ V}$.The current through the bulb is $i_1 = \frac{P}{V_b} = \frac{0.45}{1.5} = 0.3 	ext{ A}$.The voltage across the $3 \Omega$ resistor is $V_{3\Omega} = 6 	ext{ V} - 1.5 	ext{ V} = 4.5 	ext{ V}$.The total current in the circuit is $i = \frac{V_{3\Omega}}{3 \Omega} = \frac{4.5}{3} = 1.5 	ext{ A}$.The current through the resistor $R$ is $i - i_1 = 1.5 	ext{ A} - 0.3 	ext{ A} = 1.2 	ext{ A}$.Since the resistor $R$ is in parallel with the bulb,the voltage across $R$ is $1.5 	ext{ V}$.Therefore,$R = \frac{1.5 	ext{ V}}{1.2 	ext{ A}} = 1.25 \Omega$.

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) (in $\Omega$)

Similar Questions

In the circuit shown,the energy stored in the capacitor is $n\,\mu J$. The value of $n$ is ..............

Two identical fuses are rated at $10\,A$. If they are joined,

The resistance of a heater coil is $110\, \Omega$. $A$ resistance $R$ is connected in parallel with it,and the combination is joined in series with a resistance of $11\, \Omega$ to a $220\, V$ main line. The heater operates with a power of $110\, W$. The value of $R$ in $\Omega$ is

In the circuit shown in the figure,which of the following statements is incorrect?

Vedclass Test Series

Exam Paper Generator

Online Exam Module