Question

In one setup, resistors 10 Ω and 40 Ω are connected in parallel.
In another setup Resistors 30Ω, 20Ω and 60Ω are connected in parallel. These 2 are connected in series across a 12 V battery. Calculate:
• Total Resistance in the circuit
• Total current flowing in the circuit

Answer 1

Given :

In one setup

there are two resistor,

R₁ = 10Ω

R₂ = 40Ω

In another setup

there are three resistance,

R${\sf{_3}}$ = 30Ω

R${\sf{_4}}$ = 20Ω

R${\sf{_5}}$ = 60Ω

voltage of the battery, V = 12V.

To find :

(a) Total Resistance in the circuit

(b) Total current flowing in the circuit

Solution :

(a) first we have to find resistance of each set.

In 1st set,

${ \leadsto{\bf{\dfrac{1}{R'} =\dfrac{1}{R_1} + \dfrac{1}{R_2} }}}$

${ \leadsto{\bf{\dfrac{1}{R'} =\dfrac{1}{10} + \dfrac{1}{40} }}}$

${ \leadsto{\bf{\dfrac{1}{R'} =\dfrac{4 + 1}{40} }}}$

${ \leadsto{\bf{\dfrac{1}{R'} =\ \dfrac{5}{40} }}}$

${ \leadsto{\bf{\dfrac{1}{R'} =\dfrac{1}{8}}}}$

${ \leadsto{\bf{R' = 8Ω}}}$

similarly,

In another setup,

${ \leadsto{\bf{\dfrac{1}{R''} =\dfrac{1}{R_3}+ \dfrac{1}{R_4} + \dfrac{1}{R_5} }}}$

${ \leadsto{\bf{\dfrac{1}{R''} =\dfrac{1}{30}+ \dfrac{1}{20} + \dfrac{1}{60} }}}$

${ \leadsto{\bf{\dfrac{1}{R''} =\dfrac{2 + 3 + 1}{60}}}}$

${ \leadsto{\bf{\dfrac{1}{R''} =\dfrac{6}{60} }}}$

${ \leadsto{\bf{\dfrac{1}{R''} =\dfrac{1}{10} }}}$

${ \leadsto{\bf{R''= 10Ω}}}$

According to Question these are connected in series,

so total resistance in the circuit,

${\leadsto{\bf{R_s = R' +R'' }}}$

${\leadsto{\bf{R_s = 8Ω + 10Ω}}}$

${\leadsto{ \boxed{\bf{R_s = 18Ω }}}}$

thus, total resistance of the circuit is 18 Ω.

(b) using ohm's law

The potential difference V, across the ends of a given metallic wire in an electric circuit is directly proportional to the current flowing through it provided temperature remains the same .i.e.,

• V = IR

here,

V denotes potential difference

I denotes current

R denotes resistance

by substituting all the given values in the formula,

${\leadsto{\bf{I = \dfrac{V}{R} }}}$

${\leadsto{\bf{I = \dfrac{12}{18} }}}$

${\leadsto{ \boxed{\bf{I = 0.67A }}}}$

thus, Total current flowing in the circuit is 0.67A.

Remember !

SI unit of Resistance is ohms (Ω)

SI unit of current is Ampere (A)

Answer 2

$\huge {\underline{\green{Given:-}}}$

In 1 Setup

• Resistance of Resistor, R1 = 10Ω

• Resistance of Resistor ,R2 = 40Ω

In 2nd Setup

• Resistance of Resistor , R1 = 30Ω

• Resistance of Resistor ,R2 = 20 Ω

• Resistance of Resistor ,R3 = 60 Ω

• Voltage , V = 12v

$\huge {\underline{\pink{To Find:-}}}$

• Total Resistance in the circuit
• Total current flowing in the circuit

$\huge {\underline{\red{Solution:-}}}$

In 1 Setup

As we know that Equivalent Resistance in Parallel circuit is the sum of reciprocal of its individual resistance of each resistors.

• 1/Req = 1/R1 + 1/R2

Substitute the value we get

→ 1/Req = 1/10 + 1/40

→ 1/Req = 4+1/40

→ 1/Req = 5/40

→ 1/Req = 1/8

→ Req (in parallel ) = 8 Ω

Therefore the Equivalent resistance in one setup is 8 ohm.

Now, In 2nd Setup

• 1/Req' = 1/R1 + 1/R2 + 1/R3

Substitute the value we get

→ 1/Req' = 1/30 + 1/20 + 1/60

→ 1/Req' = /2+3+1/60

→ 1/Req' = 6/60

→ 1/Req' = 1/10

→ Req = 10Ω

Therefore, the Equivalent resistance in 2nd Setup is 10 ohm.

It is given that the resistance of setup 1 & 2 are connected in series .

Therefore the total resistance in the circuit is

•Req (in series) = Req(in 1st Setup) + Req' (in 2nd setup)

Substitute the value we get

→ Req = 10 + 8

→ Req = 18Ω

Total resistance in the circuit is 18 ohms.

Now ,Calculating the Current . So by using Ohm's Law

• V = IR

Substitute the value we get

→ 12 = I × 18

→ I = 12/18

→ I = 2/3 A or 0.66 A

Therefore, the current flowing through the circuit is 2/3 Ampere.