Question

The potential difference across a lamp is 12 V .How many joules of electrical energy are changed in to heat and light when i) charge of 1C passes through it (2m) 2.potential difference between 2 point of wire carrying 2 amperage current is 0.1V .calculate the resistance between these points (2m) 3.Calculate the equivalent resistance when 2 resistance of 3ohm and 6 ohm are connected in parallel(2m) 4.what will be the current drawn by the electric bulb of 40 W when it is connected to a source of 220V *​

Answer 1

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1. Heat and Light Energy

Here,

$\to V = 12 V, Q = 1C$

Work Done(W) is given by

$\to W = VQ$

$\to W = 12V\times 1C$

$\to W = 12J$

Thus, 12J of electrical energy is converted into heat and light energy.

2. Resistance of Wire

Here,

$\to V = 0.1 V, I = 2A$

By Ohm's Law, Resistance(R) is given by

$\to R = \frac{V}{I}$

$\to R= \frac{0.1V}{2A}$

$\to R = 0.05Ω$

Thus, resistance of the wire is 0.05 Ω.

3. Equivalent Resistance

Here,

$\to R_{1} = 3Ω, R_{2} = 6Ω$

Τhen equivalent resistence when they are connected in parallel is given by

$\to R_{eq} = \frac{R_{1}R_{2}}{R_{1} +R_{2}}$

$\to R_{eq} = \frac{3\times 6}{3+6}$

$\to R_{eq} =\frac{18}{9}$

$\to R_{eq} = 2Ω$

Thus,equivalent resistance is 2 Ω.

4. Current Drawn

Here,

$\to V = 220 V, P = 40W$

Then current drawn is given by

$\to I = \frac{P}{V}$

$\to I= \frac{40W}{220V}$

$\to I = 0.182A$

Thus, the electric bulb draws 0.182A of current.