Question

Two filament lamps A and B take 0.8 A and 0.9 A respectively when connected across

110 V supply. Calculate the value of current when they are connected in series across

a 220 V supply, assuming the filament resistances to remain unaltered. Also find the

voltage across each lamp​

Answer 1

Answer:

its supposed to be 220v

Explanation:

Answer 2

Given:

Current taken by lamp A = 0.8 A

Current taken by lamp B = 0.9 A

Initial Voltage across the connection = 110 V

New voltage across the connection = 220 V

To find:

Value of current in the new voltage.

The voltage across each lamp.

Solution:

In the first case, we have two different values of current, which means the connection is parallel.

We will use this relation to find the values of resistances of lamps A and B.

We know that,

V = IR

Therefore,

Resistance of lamp A (R1) = $\frac{V}{I_{1} }$

R1 = 110/0.8 = 137.5 ohm

Similarly,

Resistance of lamp B (R2) = $\frac{V}{I_{2} }$

R2 = 110/ 0.9 = 122.2 ohm

Now, It is given that we have a series connection in the second case.

therefore,

R equivalent = R1 + R2 = 137.5 + 122.2 = 259.72 ohm.

Current will be equal to

I = $\frac{V}{R_{eq} }$

I = $\frac{220}{259.72}$

I = 0.84 A

Therefore the value of the current will be 0.84 A.

Now,

The voltage across lamp A (V1) = I x R1

V1 = 0.84 x 137.5

V1 = 115.5 V

Similarly,

The voltage across lamp B (V2) = I x R2

V2 = 0.84 x 122.2

V2 = 102.6 V

Therefore, the voltage across lamp A will be 115.5 V, voltage across lamp B will be 102.6 V.