Question

Three resistances of 50, 102 and 1522 are connected in series across a supply of
240 V Find the total resistance, current drawn from the supply and voltage across
each resistance
(Ans : 300. 8A, 40 V, 80 V, 120 V]
Two resistances 252 and 5012 are connected in series across a d.c. voltage
supply. If the voltage drop across 25 resistance being 100 V. Find : (i) total
resistance (ii) current drawn from the supply and (iii) total supply voltage.
[Ans : 75. 4 A, 300 V)
Three resistances of each 3002 are connected in parallel. Determine the effective
resistance of the combination. Also find the current drawn from the supply and each
resistance when a voltage of 100 V is connected across the circuit.
(Ans: 100. 10 A, 3.33 Al​

Answer 1

Question :

• Q.1 Three resistances of 5 Ω, 10 Ω and 15 Ω are connected in series across a supply of 240 V. Find the total resistance, current drawn from the supply and voltage across each resistance.

• Q.2 Two resistances 25 Ω and 50 Ω are connected in series across a d.c. voltage supply. If the voltage drop across 25 Ω resistance being 100 V. Find :

⠀⠀⠀⠀⠀⠀(i) Total resistance in the circuit.

⠀⠀⠀⠀⠀⠀(ii) Current in the circuit.

⠀⠀⠀ ⠀ ⠀(iii) Total voltage in the circuit.

• Q3. Three resistances of each 30 Ω are connected in parallel. Determine the effective resistance of the combination. Also find the current drawn from the supply and each resistance when a voltage of 100 V is connected across the circuit.

Answer :

Q.1)

• Total resistance in the circuit = 30 Ω.
• Current drawn from the circuit = 8 A.
• Voltage across 5 Ω resistor = 40 V.
• Voltage across 5 Ω resistor = 80 V.
• Voltage across 5 Ω resistor = 120 V.

Q.2)

• Total resistance in the circuit is 75 Ω.
• Current in the circuit is 4 A.
• Total voltage in the circuit is 300 V.

Q.3)

• Effective resistance in the circuit is 100 Ω.
• Current in the Circuit is 10 A.

Given :

Q.1)

• Resistance in the resistor (R1) = 5 Ω
• Resistance in the resistor (R2) = 10 Ω
• Resistance in the resistor (R3) = 15 Ω

Q.2)

• Resistance in the resistor (R1) = 25 Ω
• Resistance in the resistor (R2) = 50 Ω
• Voltage across 25 Ω resistor, V = 100 V

Q.3)

• Resistance in the three resistors, R = 30 Ω
• Voltage through each resistors, V = 100 V

To find :

Q.1)

• Total resistance in the circuit, Rn = ?
• Current drawn from the circuit, I = ?
• Voltage across 5 Ω resistor, V(R1) = ?
• Voltage across 5 Ω resistor, V(R2) = ?
• Voltage across 5 Ω resistor, V(R3) = ?

Q.2)

• Total resistance in the circuit is, Rn = ?
• Current in the circuit, A = ?
• Total voltage in the circuit, V = ?

Q.3)

• Effective resistance in the circuit, Rn = ?
• Current in each resistance with the voltage 100 V, I' = ?

Knowledge required :

• Formula for equivalent resistance in a parallel circuit :

⠀⠀⠀⠀⠀⠀⠀1/Rn = 1/R1 + 1/R2 + 1/R3 + ... + 1/Rn

[Where, Rn = Equivalent resistance; R = Resistance in the resistor]

• Formula for equivalent resistance in a series circuit :

⠀⠀⠀⠀⠀⠀⠀Rn = R1 + R2 + R3 + ... + Rn

[Where, Rn = Equivalent resistance; R = Resistance in the resistor]

• Equation for Ohm's law :

⠀⠀⠀⠀⠀⠀⠀V = IR

[Where, V = Voltage; I = Current; R = Resistance]

Solution :

Q.1)

To find the equivalent resistance in the circuit :

By using the equation for equivalent resistance in a series Circuit and substituting the values in it, we get :

⠀⠀=> Rn = R1 + R2 + R3

⠀⠀=> Rn = 5 + 10 + 15

⠀⠀=> Rn = 30 Ω

⠀⠀⠀⠀⠀⠀∴ Equivalent resistance = 30 Ω

To find the current drawn in the circuit :

By using the ohm's law and Substituting the values in it, we get :

⠀⠀=> V = IR

⠀⠀=> 240 = I × 30

⠀⠀=> 240/30 = I

⠀⠀=> 8 = I

⠀⠀⠀⠀⠀⠀∴ Current in the circuit = 8 A

To find the voltage in across the three resistors :

For 5 Ω resistance :

⠀⠀=> V = IR

⠀⠀=> V = 8 × 5

⠀⠀=> V = 40

⠀⠀⠀⠀⠀⠀⠀∴ Voltage in 5 Ω resistor = 40 A

For 10 Ω resistance :

⠀⠀=> V = IR

⠀⠀=> V = 8 × 10

⠀⠀=> V = 80

⠀⠀⠀⠀⠀⠀⠀∴ Voltage in 5 Ω resistor = 80 A

For 15 Ω resistance :

⠀⠀=> V = IR

⠀⠀=> V = 8 × 15

⠀⠀=> V = 120

⠀⠀⠀⠀⠀⠀⠀∴ Voltage in 5 Ω resistor = 120 A

Q.2)

To find the total resistance in the circuit :

By using the equation for equivalent resistance in a series Circuit and substituting the values in it, we get :

⠀⠀=> Rn = 1/R1 + R2

⠀⠀=> Rn = 25 + 50

⠀⠀=> Rn = 75

⠀⠀⠀⠀⠀⠀∴ Equivalent resistance = 75 Ω

To find the current in the circuit :

By using the ohm's law and Substituting the values in it, we get :

⠀⠀=> V = IR

⠀⠀=> 100 = I × 25

⠀⠀=> 100/25 = I

⠀⠀=> 4 = I

⠀⠀⠀⠀⠀⠀∴ Current in the circuit = 4 A

To find the total voltage in the circuit :

By using the ohm's law and Substituting the values in it, we get :

⠀⠀=> V = IR

⠀⠀=> V = 4 × 75

⠀⠀=> V = 300

⠀⠀⠀⠀⠀⠀⠀∴ Voltage in 5 Ω resistor = 300 A

Q.3)

To find the total resistance in the circuit :

By using the equation for equivalent resistance in a parallel Circuit and substituting the values in it, we get :

⠀⠀=> 1/Rn = 1/R1 + 1/R2 + 1/R3

⠀⠀=> 1/Rn = 1/30 + 1/30 + 1/30

⠀⠀=> 1/Rn = (1 + 1 + 1)/30

⠀⠀=> 1/Rn = 3/30

⠀⠀=> Rn = 30/3

⠀⠀=> Rn = 10

⠀⠀⠀⠀⠀⠀∴ Equivalent resistance = 10 Ω

To find the current in the circuit :

By using the ohm's law and Substituting the values in it, we get :

⠀⠀=> V = IR

⠀⠀=> 100 = I × 10

⠀⠀=> 100/10 = I

⠀⠀=> 10 = I

⠀⠀⠀⠀⠀⠀∴ Current in the circuit = 10 A

Answer 2

Quêstioñ

Three resistances of 50, 102 and 1522 are connected in series across a supply of

240 V Find the total resistance, current drawn from the supply and voltage across

each resistance

(Ans : 300. 8A, 40 V, 80 V, 120 V]

Two resistances 252 and 5012 are connected in series across a d.c. voltage

supply. If the voltage drop across 25 resistance being 100 V. Find : (i) total

resistance (ii) current drawn from the supply and (iii) total supply voltage.

[Ans : 75. 4 A, 300 V)

Three resistances of each 3002 are connected in parallel. Determine the effective

resistance of the combination. Also find the current drawn from the supply and each

resistance when a voltage of 100 V is connected across the circuit.

(Ans: 100. 10 A, 3.33 Al

Answer

Rn=R1+R2+R3+....+Rn

V=IR

V=voltage

I=current

R=Resistance

Solution

Rn=R1+R2+R3

Rn=5+10+15

Rn=30Ω

equivalent resistance=30Ω

V=IR

240=I×30

240/30=I

8=I

:• current in the circuit=8A

V=IR

V=8×5

V=40

:•voltage is 5Ω resistance=40

V=IR

V=8×10

V=80

:•voltage is 5Ω resistance =80

1/Rn=1/R1+1/R2

1/Rn=1/25+1/50

1/Rn=(2+1)/50

1/Rn=3/50

Rn=16.67Ω

V=IR

10=I×10

100/10=I

:•current in the circuit=10A