Question

A resistor has a resistance of 176 ohms. How many of these resistors should be connected in parallel so that theircombination draws a current of 5 amperes from a 220 volt supply line ?
olentric boator which is connected to a 220 V supply line?​

Answer 1

Answer:

4 resistance of 176 ohm in parallel. for 5 amp from 220 v source

Explanation:

from given data of voltage and current

V=I×R

R=V/I

R=220/5

R=44 Ohm.

let suppose x no. of resistance of 176ohm connected kn parallel.

equivalent resistance will be 176/x

desire value is 44 ohm

44=176/x

x=4

4 resistance of 176 ohm in parallel.

Answer 2

AnsWer :

4 times.

Concepts Required :

• Parallel Combination - When Resistance of path change its direction, it said to be Parallel Combination.
• Req - 1/R1 + 1/R2 + 1/R3 ...... 1/Rn.
• V = IR ( Ohm's Law )

Given :

• Resistance be 176 ohm
• Current ( I ) be 5A.
• Voltage ( V ) be 220v.

Solution :

Let N times Resistors should be connected in Parallel,

We have Formula for Parallel Combination,

$\tt \dashrightarrow \frac{1}{R_{eq} }= \frac{1}{R_1 }+ \frac{1}{R_2} + \frac{1}{R_3} ..... \frac{1}{R_n}.$

• R = 176 ohm.

$\tt \dashrightarrow \frac{1}{R_{eq} }= \frac{1}{176 } + \frac{1}{176} + \frac{1}{176 } ... {n}^{th}$

$\tt \dashrightarrow \frac{1}{R_{eq} } = \frac{n}{176}$

$\tt \dashrightarrow R_{eq} = \frac{176}{n}$

$\rule{200}3$

Now, As we know Ohm's Law.

$\tt V = IR.$

Here,

• V Volt
• I Current.
• R Resistance.

We have Value, by above.

• V = 220v
• I = 5A.
• R = 176/n.

Putting all value.

$\tt \dashrightarrow 220 = 5 \times \frac{176}{n}$

$\tt \dashrightarrow n = \frac{5 \times 176}{220}$

$\tt \dashrightarrow n = 4.$

Therefore, When need to join 4 resistor in parallel.