Question

if two resistors of resistance 5watt and 10watt are connected to a battery of potential difference 6V then determine the value of current when they are consumed in a. series b. parallel ​

Answer 1

Required Answer:-

In Series:

• The resistors are 5 ohms and 10 ohms.
• Potential difference = 6 V

Equivalent resistance:

= R1 + R2

= 5 ohms + 10 ohms

= 15 ohms

Now by using ohm's law, Finding the current flowing through the wire:

⇒ V = IR

⇒ 6V = I × 15 ohms

⇒ I = 0.4 ohms (♔)

In Parallel:

• The resistors are 5 ohms and 10 ohms.
• Potential difference = 6 V

Equivalent resistance:

= R1.R2 / R1 + R2

= 5 ohms × 10 ohms / 5 ohms + 10 ohms

= 10/3 ohms

Now by using ohm's law, Finding the current flowing through the wire$:$

⇒ V = IR

⇒ 6V = I × 10/3 ohms

⇒ I = 1.8 ohms (♔)

Answer 2

Correct Question :

• If two resistors of resistance 5 Ω and 10 Ω are connected to a battery of potential difference 6V then determine the value of current when they are consumed in a. series b. parallel ​

Given :

• R = 5 Ω
• R' = 10Ω
• Voltage = V= 6 V

To find :

Current flowing through the circuit when resistors are connected in

• Series
• Parallel

Solution :

(a) Series

Equivalent resistance of the combination when two resistors are connected in series

⇒ Rs = R + R'

⇒ Rs = 5 + 10

⇒ Rs = 15 Ω

Now let's find the current flowing through the circuit

By using ohm's law ,

⇒ I = V/ Rs

⇒ I = 6 / 15

⇒ I = 0.4 A

The current flowing through the circuit when both resistors are connected in series is 0.4 A .

(b) Parallel

Equivalent resistance of the combination when two resistors are connected in parallel

⇒ 1 /Rp = 1/ R + 1 / R'

⇒ 1 / Rp = 1 / 5 + 1 / 10

⇒ 1 / Rp = 2 + 1 / 10

⇒ 1 / Rp = 3 / 10

⇒ Rp = 10 / 3

⇒ Rp = 3.3 Ω( approx.)

Now let's find the current flowing through the circuit

By using ohm's law ,

⇒ I' = V/ Rp

⇒ I' = 6 / 3.3

⇒ I '= 1.8  A (approx.)

The current flowing through the circuit when both resistors are connected in parallel is 1.8 A .