Question

Find the equivalent resistance of the following circuit, and hence find the total current

drawn by the circuit from the battery.​

Answer 1

• Total Resistance = 10/3 ohm
• Current drawn by the circuit is 3 Ampere.

Given:-

• Resistance of Each Resistors = 100 ohm

To Find:-

• Equivalent Resistance ,Req
• Current drawn by the circuit ,I

Solution:-

As we have to calculate the Equivalent resistance of the given circuit.

Let's arrange the resistors which is given in the circuit in favourable way. The new diagram formed are given in attachment .

Calculating the Equivalent Resistance or Total Resistance

As we know that the Equivalent Resistance is defined as the sum of reciprocal of Resistance of each individual Resistors.

• 1/Req = 1/R1 + 1/R2 + 1/R3

Substitute the value we get

→ 1/Req = 1/10 + 1/10 + 1/10

→ 1/Req = 1+1+1/10

→ 1/Req = 3/10

→ Req = 10/3 Ω

• Hence, the total Resistance across the given Resistor is 10/3 ohms.

Now, Calculating the current drawn by the circuit .

Using Ohm's Law

• V = IR

where,

V denote Potential Difference

I denote Current

R denote Resistance

Substitute the value we get

→ 10 = I × 10/3

→ I = (10/3)/10

→ I = 3 A

• Hence, the current flow through the circuit is 3 Ampere.

For More Information Refer to Attachment !!

Answer 2

$\bf\blue{As\:shown\:in\:the\:attachment\:diagram,}$

➣ R₁ , R₂ & R₃ are connected in parallel combination.

• R₁ = 10 Ω

• R₂ = 10 Ω

• R₃ = 10 Ω

$\bf\purple{As\:we\:know\:that,}$

☆ Equivalent resistance in parallel combination is,

$\red\bigstar\:\:{\underline{\green{\boxed{\bf{\color{peru}\dfrac{1}{R_{eq}}\:=\:\dfrac{1}{R_1}\:+\:\dfrac{1}{R_2}\:+\:\dfrac{1}{R_3}\:}}}}}$

$:\implies\:\:\bf{\dfrac{1}{R_{eq}}\:=\:\dfrac{1}{10}\:+\:\dfrac{1}{10}\:+\:\dfrac{1}{10}\:}$

$:\implies\:\:\bf{\dfrac{1}{R_{eq}}\:=\:\dfrac{1\:+\:1\:+\:1}{10}\:}$

$:\implies\:\:\bf{\dfrac{1}{R_{eq}}\:=\:\dfrac{3}{10}\:}$

$:\implies\:\:\bf{R_{eq}\:=\:\dfrac{10}{3}\:}$

$:\implies\:\:\bf\orange{R_{eq}\:=\:3.34\:\Omega}$

$\Large\bold\therefore$ The equivalent resistance of the circuit is 3.34 Ω.

$\bf\pink{Again\:we\:know\:that,}$

$\purple\bigstar\:\:{\underline{\blue{\boxed{\bf{\color{olive}Current\:(I)\:=\:\dfrac{Voltage}{R_{eq}}\:}}}}}$

$\bf\red{Where,}$

• Voltage (V) = 10 V

• $\rm{R_{eq}}$ = 3.34 Ω

$:\implies\:\:\bf{I\:=\:\dfrac{10}{\frac{10}{3}}\:}$

$:\implies\:\:\bf{I\:=\:10\times{\dfrac{3}{10}}\:}$

$:\implies\:\:\bf\green{I\:=\:3\:A}$

$\Large\bold\therefore$ The total current flowing through the circuit is 3 A.