Question

Two lamps rated 60W; 240V and 100W, 240Vrespectively are connected in series to a 240V power source. Calculate;
a) The resistance of each lamp.
b) The effective resistance of the circuit.
c) The current passing the lamps. pls answer correctly ​

Answer 1

Answer:

See the answers below.

Explanation:

The total power of the circuit is equal to the sum of the powers of each lamp.

$P=60+100\\P=160 [W]$

Now we have a voltage source equal to 240 [V], so by means of the following equation we can find the current circulating in the circuit.

$P=V*I$

where:

P = power [W]

V = voltage [V]

I = current [amp]

$I = P/V\\I=160/240\\I=0.67 [amp]$

So this is the answer for c) I = 0.67 [amp]

We know that the voltage of each lamp is 240 [V]. Therefore using ohm's law which is equal to the product of resistance by current we can find the voltage of each lamp.

a)

$V=I*R$

where:

V = voltage [V]

I = current [amp]

R = resistance [ohms]

Therefore we replace this equation in the first to have the current as a function of the resistance and not the voltage.

$P=V*I\\and\\V = I*R\\P = (I*R)*I\\P=I^{2}*R$

$60 = (0.67)^{2}*R\\R_{60}=133.66[ohm] \\and\\100=(0.67)^{2} *R\\R_{100}=100/(0.66^{2} )\\R_{100}=225 [ohm]$

b)

The effective resistance of a series circuit is equal to the sum of the resistors connected in series.

$R = 133.66 + 225\\R = 358.67 [ohms]$