Question

A parallel combination of three resistors takes a current of 7.5A from a 30V battery. if two of the resistances are 12 ohm and 10 ohm, find the third one?

Answer 1

Equivalent resistance = V / i
= 30 V / (7.5 A)
= 4 Ω

In parellel arrangement
1/Req = 1/R₁ + 1/R₂ + 1/R₃
1/R₃ = 1/Req - (1/R₁ + 1/R₂)
1/R₃ = 1/4 - (1/10 + 1/12)
R₃ = [1/4 - (1/10 + 1/12)]^-1
R₃ = 15 Ω

Resistance of third resistor is 15 Ω

Answer 2

Answer:

The third one of the resistance is      $R=15 \Omega$

Explanation:

Given:

Two of the resistances are 12 ohm and 10 ohm

Three resistors take a current of 7.5A

Current = 30V

To find,

The third one of the resistance

A resistor is an acquiescent two-terminal electrical element that executes electrical resistance as a circuit element. Resistors decrease the current flow and lower voltage levels within circuits. Most circuits usually have better additional than one resistor to determine the flow of charges in a circuit. The two easiest combinations of resistors are – series and parallel.

$\left.\right|_{\text {Total }}=7.5 \mathrm{~A}$

$V=30 \mathrm{~V}$

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

$i.e. R_{\text {Total }}=4 \Omega$

Let the third resistor be $\mathrm{R}$.

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

$\therefore(1 / R)=(1 / 4)-(1 / 10)-(1 / 12)$

$\therefore(1 / R)=\{(15-6-5) /(60)\}$

$\therefore(1 / R)=(4 / 60)=(1 / 15)$

$R=15 \Omega$

Hence, The third one is $R=15 \Omega$

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