Question

Three resistors of 4Ω, 6Ω and 8Ω are connected in parallel supplied with a battery of 12V.

a) Draw the circuit diagram with an ammeter and a voltmeter.

b) Find current through each resistor.



c) Find the total resistance of the circuit.

d) Find the total current in the circuit.​

Answer 1

Answer:

سہس رحمان نے لکھا ۳تبصرے اردو غزل میں نے لکھا ہے کہ اس ۳سالہ پرانے وقتوں میں ایک دوسرے سے رابطہ کرتے ہوئے اس کی سخت مخالفت کرتے ہیں تاکہ اس سے زیادہ ہے اور مردوں کی اوسط عمر کے بچوں کا معاملہ میں ایک ہی ہے تو پھر وہ گھر بیٹھے ہیں کہ ان کا تعلق لاہور سے تھا

Answer 2

Given :

• $\sf R_1 = 4Ω$
• $\sf R_2 = 6 Ω$
• $\sf R_1 = 8Ω$
• $\sf Volt (v) = 12 v$

To Find :

a) Draw the circuit diagram with an ammeter and a voltmeter.

b) Find current through each resistor.

c) Find the total resistance of the circuit.

d) Find the total current in the circuit.

Formulas Applied :

Case 1 :-

Current through each resistor :-

$\underline{\boxed{\sf I = I_1 + I _2 + I_3}}$

In which ,

• $\sf I_1 = \dfrac{V}{R_1}$

• $\sf I_1 = \dfrac{V}{R_1}$

• $\sf I_1 = \dfrac{V}{R_1}$

Case 2 :-

Equivalent Resistance of 3 resistors in parallel circuit is :-

$\underline{\boxed{\sf \dfrac{1}{R_p} = \dfrac{1}{R_1}+ \dfrac{1}{R_2} +\dfrac{1}{R_3}}}$

Case 3 :-

According to ohm's law,

$\underline{\boxed{\sf I = \dfrac{V}{R}}}$

In which ,

• I = Total current
• R = Total effective Resistance
• V = Volt

Solution :

b) Current through each resistor

$\implies \sf I_1 = \dfrac{V}{R_1}\Rightarrow \dfrac{12}{4}\Rightarrow \underline{3\:ampere}$

$\implies \sf I_2= \dfrac{V}{R_2}\Rightarrow \dfrac{12}{6}\Rightarrow \underline{2 \:ampere}$

$\implies \sf I_3 = \dfrac{V}{R_3}\Rightarrow \dfrac{12}{8}\Rightarrow \underline{1.5\:ampere}$

c) Total Resistance in the circuit

$\implies \sf \dfrac{1}{R_p} = \dfrac{1}{R_1}+ \dfrac{1}{R_2} +\dfrac{1}{R_3}$

$\implies \sf \dfrac{1}{R_p} = \dfrac{1}{4}+ \dfrac{1}{6} +\dfrac{1}{8}$

$\implies\sf \dfrac{1}{R_p} = \dfrac{1\times 6}{4\times 6}+ \dfrac{1\times 4}{6\times 4} +\dfrac{1\times 3}{8\times 3}$

$\implies\sf \dfrac{1}{R_p} = \dfrac{6}{24}+ \dfrac{4}{24} +\dfrac{3}{24}$

$\implies\sf \dfrac{1}{R_p} = \dfrac{6+4+3}{24}\Rightarrow \dfrac{13}{24}$

$\implies\sf R = \dfrac{24}{13}\Rightarrow \boxed{\sf R_p = 1.8 Ω}$

d) Total Current in the circuit

$\implies \sf I = \dfrac{V}{R}$

$\implies \sf I = \dfrac{12}{1.8}$

$\implies \sf I = 6\: ampere$

Required Answer :

b)Current through each resistors is $\sf\underline{3\:ampere}, \underline{2\:ampere}\:and\:\underline{1.5\:ampere}$ Respectively.

c)Total Resistance in the circuit is$\underline{\sf R_p = 1.8 Ω}$

d)Total current in the circuit is $\underline{ \sf 6\: ampere}$