A resistor has a resistance fo 176 ohms. How many of these resistors should be connected in parallel so that their combination draws a current of 5 amperes from a 220 volt supply line?

Open in App

Solution

For Resistors in Parallel,

$\frac{1}{R_{t o t a l}} = \frac{1}{R_{1}} + \frac{1}{R_{2}} + \frac{1}{R_{x}}$

From Ohm's Law

V = IR
220 = (5)R
R = 44 Ω
$∴ \frac{1}{44} = N o . o f 176 Ω R e s i s t o r s x \frac{1}{176}$

Let no. of 176Ω Resistors be x

Then $\frac{1}{44} = \frac{x}{176}$

$x = 4$

$∴$ 4 resistors of 176Ω ( in parallel) are required to carry 5 A on a 220 V line.

Suggest Corrections

Similar questions

Q. How many resistors of 160 Ω should be connected in parallel to get 11 A current from a 220 V supply?

Q. A student was given three resistors of 2 ohms, 3 ohms and 5 ohms by his teacher. The teacher asked him to connect these three resistors in such a way so as to obtain the minimum resistance.
(a) In what way should the student connect the resistors?
(b) Draw a diagram of the combination of resistors.

Q. Two resistances 6 and 12 ohms are connected in parallel to a battery of 10 volts. The current in amperes in 12 ohm resistor is

Two resistors A and B of resistance 4 ohm and 6 ohm respectively are connected in parallel. The combination is connected across a 6 volt battery of negligible resistance. Calculate: (i) the power supplied by the battery, and (ii) the power dissipated in each resistor.

Join BYJU'S Learning Program

A resistor has a resistance fo 176 ohms. How many of these resistors should be connected in parallel so that their combination draws a current of 5 amperes from a 220 volt supply line?

For Resistors in Parallel, 1Rtotal=1R1+1R2+1Rx From Ohm's Law V = IR 220 = (5)R R = 44 Ω ∴ 144=No.of176ΩResistorsx1176 Let no. of 176Ω Resistors be x Then 144=x176 x=4 ∴ 4 resistors of 176Ω ( in parallel) are required to carry 5 A on a 220 V line.