Question

alterence and resistance.
(c)
How many 132 resistors in parallel are required to carry 5 A on a
220 V line ?​

Answer 1

Answer:

Three

Explanation:

Given :

• Many 132 ohms resistors are connected in parallel combination

• A 220 volts source produces a potential difference of 220 volts

• A current of 5 Amperes is produced in the circuit

To find :

• Number of 132 ohms resistors to be connected in parallel

Using the ohms law :

V=I×R

220=5×R

R=220/5

R=44 ohms

When resistors of same resistance are connected in parallel, equivalent resistance = resistance /number of resistors

Let number of resistors = X

132/X=44

44X = 132

X=132/44

X=3

Three 132 ohms resistors must be connected in parallel for the above combination

Answer 2

$\bigstar$ $\sf{\purple{\underline{\underline{\blue{To\:Find:-}}}}}$

• How many “132 Ohm” resistors in parallel are required ?

$\bigstar$ $\sf{\blue{\underline{\underline{\purple{SOLUTION:-}}}}}$

☞ GIVEN :-

• Current (I) = 5 A

• Voltage (V) = 220 V

• resistance of resistor (R) = 132 Ohm

☞ We have know that

• when “n” number of resistors are connected in parallel , then the Equivalent Resistance is “R/n” .

☞ So, Here

$\tt{\blue{\implies\:{\dfrac{R}{n}}\:=\:{\dfrac{132}{n}}\:}}$

☞ FORMULA :-

$\tt{\green{\boxed{\underline{Voltage\:(V)\:=\:Current\:(I)\:\times\:Resistance\:(R)\:}}}}$

☞ CALCULATION :-

$\tt{\implies\:V\:=\:I\:\times\:{\dfrac{R}{n}}\:}$

$\tt{\implies\:220\:=\:5\:\times\:{\dfrac{132}{n}}\:}$

$\tt{\implies\:n\:=\:{\dfrac{5\times132}{220}}\:}$

$\tt{\implies\:n\:=\:3\:}$

☞ Therefore , There are 3 “132 Ohm” resistors in parallel are required to carry 5A current on a 220V line .

$\bigstar\:\underline{\boxed{\bf{\red{Required\:Answer\::\:3\:}}}}$