Question

in a given circuit find out the current through 18 ohm resistor and the net resistance of the circuit​

Answer 1

resistance 6 & 12 ohm are paralle and thats value will be 4 ohm which will be series with 8 ohm resistor thats 12 ohm. This will be parallel to the other 12 ohm resistor Its value is 6ohm.

Therefore net resistance = 18 + 6=> 24 ohm.

since the final two resistance are in series combination value of current remains constant through out the circuit.

Thus I = V/R =>24/24

I = 1 A

Hope its helpful!

Answer 2

Answer:

The current through 18 ohm resistor is 1 Ampere and net resistance of the circuit is 24 ohms.

Explanation:

• The resistances between point a and b is connected is parallel.
• Let, the effective resistance between points a and b is R(ab),

$\frac{1}{R(ab)} = \frac{1}{6} + \frac{1}{12} \\ = \frac{2 + 1}{12} \\ = \frac{3}{12} \\ \frac{1}{R(ab)}= \frac{1}{4} \\ R(ab) = 4 \: ohms$

• Resistance, R(ab) and 8 ohm resistance is in series.
• Net effective resistance in upper arm of cd path is (8 + 4)ohms = 12 ohms.
• The effective resistance between points c and d is

$\frac{1}{R(cd)} = \frac{1}{12} + \frac{1}{12} \\ = \frac{2}{12} \\ = \frac{1}{6} \\ R(cd) = 6 \: ohms$

• Now, resistance R(cd) and 18 ohm are connected in series.
• Effective resistance of circuit is (18 + 6) = 24 ohms.
• Current through the circuit is given by

$current = \frac{voltage}{resistance} \\ i = \frac{24}{24} \\ i = 1A$

• Since, 18 ohm resistor is connected in series with battery. Current through 18 ohm resistor is 1 Ampere.
• Hence, effective resistance of circuit is 24 ohms and current through 18 ohm resistor is 1 Ampere.