Question

Draw a schematic diagram of a circuit consisting of a batter/ of four 2 V cells. A 5 ohm resister, an 8 ohm resisters, a 12 ohm resister and a plug key, all connected In series 5. calculate (i) Total Resistance (ii) Total current​

Answer 1

Explanation:

8V battery by mistake it 6V

Total resistance = 8 +5+12= 25 ohms

total current V = IR

BUT 8= I × 25

I = 8/25

I = 0.32A

Answer 2

Answer:

The total resistance of the circuit is 25 Ω

The current in the circuit is 0.32 A

Explanation:

Refer the attachment for the diagram.

There are 4 cells of 2 V, so the battery will be of,

$\rightarrow$ 2 × 4

$\rightarrow$ 8 V

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• i) Total Resistance :

The given resistors are in series connection.

$\sf{\rightarrow} \: R_s = R_1 + R_2 + R_3$

$\sf{\rightarrow} \: R_s = 5 + 8 + 12$

$\sf{\rightarrow} \: R_s = 5 + 20$

$\sf{\rightarrow} \: R_s = 25 \:ohms$

The total resistance of the circuit is 25 Ω

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• Current :

Following the Ohm's Law,

$\sf{\rightarrow} \: Current \: (I) = \dfrac{Voltage \: (V)}{ Resistance \: (R)}$

• V = 8 V
• R = 25 Ω

$\sf{\rightarrow} \:I = \dfrac{8}{25}$

$\sf{\rightarrow} \:I = 0.32 \:A$

Current = 0.32 A

The current in the circuit is 0.32 A