Question

5. An electric bulb marked 500 W, 250 V is connected to a 250 V supply.
a. Find the electric current through the circuit.
b. Calculate the resistance of the bulb.
6. a. What is meant by amperage?
b.An appliance of power 690 W is used in a branch circuit. If the voltage is 230 V, what is its
amperage?
7. If a bulb is lit after rejoining the parts of a broken filament, what change will occur in the intensity of
the light from the lamp? What will be the change in the power of the bulb?

Please answer

Answer 1

Answer:

Explanation:

a. P=500 W,V=250 V  

   P=VI

   I=  

V

P

​

=  

250

500

​

=2 A

b. R=  

I

V

​

=  

2

250

​

=125 Ω

Answer 2

• Question 5

A) We know that,

w = I × V

• where, w = power, I = currect, v = volt

And from the question we already have,

• voltage = 250 v and power = 500 W

So, the electric current in the circuit will be;

500 = 250 × I

500/250 = l

2 = l

Therefore, the current in the circuit is 2 A.

_________________________

B) We know that,

R = Vl

• And we have calculated the current that is 2 A, the voltage is given as 250 v

R = 250/2

R = 125 ohm

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• Question 6

A) One amphere is the current in a circuit, when one coulumb of charge flows through a conductor in one second.

B) We know that,

w = I × V

• from the ques, we have; w = 690 w, voltage = 230 V

On putting thus in formula,

690 = 230 × l

690/230 = l

3 = l

Thus, the amperage is 3 A.

_________________________

• Question 7

There are possibilities that the bulb will not glow, as the filament is broken it may happen that the current will not be able to flow through it accordingly.

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