Question

An electric lamp whose resistance is 20 Ω and a conductor of 4 Ω resistance

are connected in series to a 6V battery as shown in figure. Then the total current

flowing through the circuit is​

Answer 1

Explanation:

. the total resistance of the circuit =20Ω+4Ω=24Ω

b). The current through the circuit= current through the bulb= current through the conductor.

The current through the circuit=

the voltage of the battery

voltage applied

.

the current through the circuit =

24

6

amp

The current through the circuit = 0.25amp

c). The potential difference across the lamp and the conductor.

(i) The potential difference across the electrical lamp =0.25×20=5volt

(ii) The potential difference conductor =0.25×4=1 volt

d). Power can be calculated as

current through the lamp

voltage across the lamp

power=

0.25

5

→20W.

Answer 2

Hint: First we have to calculate resistance using the series rule and using total resistance we can calculate total current in the circuit and by using value of the current we can calculate potential difference across the electrical lamp and the conductor and we can calculate power using total current in the circuit and the voltage.

Formula used:

I=VRP=VI

Complete step by step solution:

(A) The total resistance of the circuit:

As we can see in the circuit that electric lamp with the resistance 20Ω and the conductor 4Ω are connected in the series, So that we can apply series rule of resistance.

Now total resistance of the circuit:

R=20+4R=24Ω

Hence the total resistance of the circuit is 24Ω.

(b) The current through the circuit:

We know that formula for the current (I) :

⇒I=VR....(1)

Where, I = current

V = Voltage applied

R = total resistance

Here voltage is given as,

V = 6V

And the total resistance

R = 24Ω

Now current,

I=624∴I=0.25A

Hence current through the circuit is 0.25A.

(c) The potential difference across the

(i) Electric lamp.

If the potential difference across the electric lamp is Ve and the resistance of the electric lamp Re then,

⇒Ve=I×Re⇒Ve=0.25×20∴Ve=5V

Hence the potential difference across the electric lamp is 5V.

(ii) The potential difference across the conductor.

If the potential difference across the conductor is Vc and resistance of the conductor is Rc then,

⇒Vc=I×Rc⇒Vc=0.25×4∴Vc=1V

(D) Power of the lamp:

Formula for the power,

P = voltage across the lamp (Ve)current through lamp (I)

Where, P is power

⇒P=50.25∴P=20W

Note:

Here in the option (d) when we calculate power of the lamp don’t put the voltage as the total applied voltage instead of the voltage across the lamp (Ve) that we find in the option (c).