Question

draw a schematic diagram of a circuit consisting of a battery of three cell of 2v each ,a5ohm resistor an 8ohm resistor and a 12ohm resistor and a plug key all cannected in series also calculate: (a) total resistance of the circuit.(b) voltage across 8ohm resistor​

Answer 1

Total Resistance = 25 ohm

Voltage across 8 ohm resistor is 33.28 volts.

Given:-

• Potential difference ,V = 2 × 3 = 6 v
• Resistance of Resistor ,R1 = 5 ohm
• Resistance of Resistor ,R2 = 8 ohm
• Resistance of Resistor ,R3 = 12 ohm

To Find:-

• Equivalent Resistance ,Rs
• Voltage across 8 ohm resistor ,V'

Solution:-

Firstly we calculate the Equivalent resistance in Series.

Equivalent Resistance in series can be calculated by the Sum of resistance of individual resistor.

• Rs = R1 + R2 + R3

Substitute the value we get

→ Rs = 5 + 8 + 12

→ Rs = 25 ohm

• Hence, the total resistance of the circuit is 25 ohms

Now, Calculating the current flow through the circuit.

Using Ohm's Law

• V = IR

Substitute the value we get

→ 6 = I × 25

→ I = 25/6

→ I = 4.16 A

As we know that the current flow through each Resistors when they are connected in series is same .

So, the Potential Difference or Voltage across the 8 ohm resistor .

Using Ohm's Law

→ V' = 4.16 × 8

→ V' = 33.28 v

• Hence, the voltage across 8 ohm resistor is 33.28 volts.

Answer 2

Answer:

Given :-

• Three cell of 2 V = 3×2 = 6 V
• Resistors of 5 Ohm,8 Ohm, 12 Ohm

To Find :-

• Total Resistance
• Voltage accross 8 Ohm resistor

Solution :-

As we know that

Rs = R1 + R2 + R3

Rs = 5 + 8 + 12

Rs = 25 Ohm

Now,

Finding current using Ohm law

$\huge \fbox{V = IR}$

6 = I × 25

I = 25/6

I = 4.16 Ampere

Now,

Again using same method

V = 4.16 × 8

V = 33.28 V