Question

Calculate the electric current in the given circuit when i) key K is open ii) key K is closed

Answer 1

Answer:

(i) 0A

(ii) 1A

Explanation:

(i) when the key is open, electricity will not pass. Therefore, current =0A.

(ii) when key is closed,

total resistance= R =6 ohm

V=iR => 6= i 6 => I=1A.

Pls mark me as the Brainliest!! ;)

Answer 2

Answer:

i) 0.5 A

ii) 1 A

Explanation;

Taking a look at the diagram, we can see that the resistors are connected in a parallel combination. [ Two 6 ohm resistors in parallel with the other two ].

Now, we can also see that the key is only connected to the top branch of the combination.

Case 1: Key is open

If we leave the key open, current would still flow in the circuit. This is because, it takes the second branch which is below since the top branch forms an incomplete circuit.

Therefore calculating net resistance in the second branch, it comes out to be 12 ohm [ 6 + 6 ].

Applying Ohm's law, we get:

→ I = V / R

→ I = 6 / 12 = 0.5 A

Hence 0.5 A of current flows when the key is open.

Case 2: Key is closed

Now we see that the circuit is a simple combination of resistors in parallel and series. Calculating the net resistance we get:

$\dfrac{1}{R} = \dfrac{1}{6+6} + \dfrac{1}{6+6}\\\\\dfrac{1}{R} = \dfrac{1}{12} + \dfrac{1}{12}\\\\\dfrac{1}{R} = \dfrac{2}{12} = \dfrac{1}{6}\\\\\implies R = 6 \:\Omega$

Now applying Ohm's Law we get:

→ I = V / R

→ I = 6 V / 6 Ω

→ I = 1 A

Hence 1 A of current flows when the key is closed.