Question

A circuit consists of four 100W lamps

Answer 1

Complete question is;

A circuit consists of four 100-W lamps connected in parallel across a 230-V supply. Inadvertently, a voltmeter has been connected in series with the lamps. The resistance of the voltmeter is 1500 Ω and that of the lamps under the conditions stated is six times their value then burning normally. What will be the reading of the voltmeter?

Answer:

150.42 V

Explanation:

We are told that the circuit consists of four 100W lamps.

We know that Power is given by the equation;

P = V²/R

Thus;

R = V²/P

Now, we are told that the four lamps are connected in parallel across a 230V supply.

Thus, V = 230 V

So resistance, R = 230²/100

R = 529 Ω

We are told that the resistance of the lamps under the conditions stated is six times their value when burning normally.

Thus, total resistance of each lamp under the conditions = 529 × 6 = 3174 Ω

So, since they are connected in parallel, equivalent resistance for each lamp = 3174/4 = 793.5 Ω

Now, since this resistance is connected in series with the voltmeter resistance of 1500 Ω

Therefore, total circuit resistance = 1500 + 793.5 = 2293.5 Ω

Thus;

circuit current = 230/2293.5 = 0.100283 A

Now, according to Ohm’s law, voltage drop across the voltmeter = 1500 × 0.100283 ≈ 150.42V

Answer 2

CHECK THE COMPLETE QUESTION BELOW

A circuit consists of four 100-W lamps connected in parallel across a 230-V supply. Inadvertently, a voltmeter has been connected in series with the lamps. The resistance of the voltmeter is 1500 Ω and that of the lamps under the conditions stated is six times their value then burning normally. What will be the reading of the voltmeter?

Answer:

the reading of the voltmeter=150.4V

Explanation:

We can determine the wattage of a lamp using below expression:

: W = I² R....................eqn(1)

But fro ohms law V=IR

then I= V/R

If we substitute I into equation (1)

We have W= V²/R

But W= 100W

V= 230V

Then

W=220²/R

100 = 2302/R

R = 529 Ω

We can as well calculate the Resistance of each lamp under given condition that they are sixtimes their value when burning normally.

R = 6 × 529 = 3174 Ω

We can also calculate quivalent resistance of the abovefour lamps connected in parallel then

R = 3174/4

= 793.5 Ω

total circuit resistance can be calculated since we know that resistance is connected to voltmeter of 1500 Ω resistance in series arrangement

Then

total circuit resistance = 1500 + 793.5

= 2293.5 Ω

Then from ohms law again

I= V/R

circuit current = 230/2293.5 A

The reading of the voltage drop across the voltmeter

= 1500 × 230/2293.5

= 150.4V