Question

A 4Ω resistor is connected in series with a 2Ω lamp, and both are connected across battery of 3V. Find the

current flowing through the lamp and potential difference across 4Ω resistor​

Answer 1

Answer:

So, current through the lamp = 0.5 Amperes

Potential difference across 4ohm resistor =

2volts

Explanation:

Gievn :

- Resistance of resistor ( R1 ) = 4 Ohm.

- Resistance of lamp ( R2 ) = 2 ohm.

- Voltage applied ( V ) = 3 Volts.

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Now,

(i) Current throughout the circuit is same as in

lamp as they are in series connection. So,

-> V = iR

now equivalent resistance of series circuit is

-> R = R1 + R2

-> R = 2 + 4 = 6 ohms.

now by substituting it in ohms law

-> 3 = i(6)

-> i = 3/6

-> i = 0.5 Ampere.

(ii) potential diference across resistor, let it be V1.

-> V1 = i(R1)

-> V1 = 0.5(4)

-> V1 = 2 volts.