If a 30, V, 90, W bulb is to be operated on a | vedclass

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

Question

(C) The rated power of the bulb is $P = 90\, W$ at a voltage of $V_b = 30\, V$. The current $i$ flowing through the bulb when operating at its rated p

Vedclass · Accepted Answer

$30$

Vedclass · Answer

(C) The rated power of the bulb is $P = 90\, W$ at a voltage of $V_b = 30\, V$. The current $i$ flowing through the bulb when operating at its rated power is given by $i = \frac{P}{V_b} = \frac{90\, W}{30\, V} = 3\, A$.Since the bulb is connected in series with a resistor $R$ to a $120\, V$ supply,the voltage drop across the resistor must be $V_R = 120\, V - 30\, V = 90\, V$.Using Ohm's law for the resistor,$V_R = i 	imes R$,we get $90\, V = 3\, A 	imes R$.Therefore,$R = \frac{90\, V}{3\, A} = 30\, \Omega$.

If a $30\, V, 90\, W$ bulb is to be operated on a $120\, V$ line,what resistance in ohms should be connected in series with the bulb?

Similar Questions

For the given circuit,the $emf$ of the cells are $E_1 = 4 \, V$ and $E_2 = 8 \, V$,and their internal resistances are $0.5 \, \Omega$ and $1 \, \Omega$ respectively. Calculate the terminal voltage of each cell.

Twelve wires each having resistance $2 \Omega$ are joined to form a cube. $A$ battery of $6 \text{ V}$ emf is connected across points $a$ and $c$. The voltage difference between $e$ and $f$ is . . . . . . $\text{V}$.

For the network of resistors shown in the figure,choose the correct option.

In the following circuit, a power of $50 \ W$ is absorbed in the section $AB$ of the circuit. The value of resistance '$X$' is (in $\Omega$)

Two electric bulbs whose resistances are in the ratio of $1 : 2$ are connected in parallel to a constant voltage source. The powers dissipated in them have the ratio:

Vedclass Test Series

Exam Paper Generator

Online Exam Module