Question

A prallel combination registar take's a current of 7.5 Ampere from a 30 volt bettery. If the 2 registance are 10 ohm & 12ohm. Find the 3rd registance??​

Answer 1

Answer:

15

Explanation:

I= 7.5A

V = 30V

(V/I) = {(30)/(7.5)} = 4

RTotal = 4Ω Let third resistor be R.

hence (1/4) = (1/10) + (1/12) + (1/R)

∴ (1/R) = (1/4) – (1/10) – (1/12)

∴ (1/R) = {(15 – 6 – 5)/(60)}

∴ (1/R) = (4/60) = (1/15) R = 15Ω

Answer 2

$\large\boxed{\mathtt\red{Solution:-}}$

Given :-

◑ Electric current = 7.5 Ampere

◑ Potential Difference of the battery = 30 volts

◑ Resistance-1 (R₁) = 10 Ω

◑ Resistance-2 (R₂) = 12 Ω

To Find :-

◑ 3rd Resistance (R₃) of the parallel combination.

Solving :-

✏ First, find the total resistance :-

$\large\mathtt{Total\: Resistance=\frac{Potential\: Difference}{Electric\: Current}}$

✒ Total Resistance = $\large\mathtt{\frac{7.5\:A}{30\:V}}$

∴ T.R (Total Resistance) = 4 Ω

✏ Now, Applying the following formula:-

➽$\large\mathtt\blue{\:\frac{1}{TR}=\frac{1}{R₁}+\frac{1}{R₂}+\frac{1}{R₃}}$

✒ $\large\mathtt{\frac{1}{4}=\frac{1}{10}+\frac{1}{12}+\frac{1}{R₃}}$

✒$\large\mathtt{\:\frac{1}{R₃}=\frac{1}{4}-\frac{1}{10}-\frac{1}{12}}$

✒$\large\mathtt{\:\frac{1}{R₃}=\frac{15-6-5}{60}}$

✒$\large\mathtt{\:\frac{1}{R₃}=\frac{4}{60}}$

✒$\large\mathtt{R₃=\frac{60}{4}}$

$\large\boxed{\mathtt\green{∴\:R₃=15 Ω}}$