Question

7. The rating of bulb B_{1} is 60 V, 12 W & bulb B_{2} is 100 V, 100 W. Find the maximum emf of the battery so that all bulbs remain safe.​

Answer 1

Answer:

Medium

Solution

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Correct option is D)

Resistance is given

Explanation:

Medium

Solution

verified

Verified by Toppr

Correct option is D)

Resistance is given by R=

P

V

2

∴ Resistance of 100 W bulb is given by

R

1

=

100

V

2

=

100

(250)

2

Answer 2

To determine the maximum emf of the battery that ensures the safety of both bulbs, we need to consider the power rating and voltage rating of each bulb.

The power rating of bulb B1 is 12 W, and its voltage rating is 60 V. The power rating of bulb B2 is 100 W, and its voltage rating is 100 V.

We can use the formula P=V^2/R, where P is power, V is voltage, and R is resistance, to find the resistance of each bulb. Using this formula, we get:

For bulb B1: R = (V^2) / P = (60^2) / 12 = 300 ohms

For bulb B2: R = (V^2) / P = (100^2) / 100 = 100 ohms

Now, we can use the formula for the total resistance of a circuit in series, which is R_total = R1 + R2, where R1 and R2 are the resistances of bulbs B1 and B2, respectively.

R_total = R1 + R2 = 300 + 100 = 400 ohms

Finally, we can use Ohm's law, which states that V = IR, where V is voltage, I is current, and R is resistance, to find the maximum emf of the battery:

V_max = I * R_total

To ensure the safety of both bulbs, we want the current to be the same in each bulb. Therefore, we can assume that the current through both bulbs is equal to the total current in the circuit, which is I = V_max / R_total.

Substituting this expression for I in the equation for V_max, we get:

V_max = (V_max / R_total) * R_total

V_max = I * R_total

V_max

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