Question

Pls show the working and reason​

Answer 1

Answer:

8/3A

Explanation:

From the figure we can identify that

120V is across the circuit.

When you simplify the circuit you get 4ohm,2ohm and 4ohm resistance in series

So total resistance is 10ohm

And the current flowing through the circuit using ohm's law

V=IR

120 = I * 10

I = 12A (in fig 5 the current is 12A)

We need to calculate the voltage across each resistance in series in fig. 5

Using ohms law again

Across 4ohm, V = IR = 12*4 = 48V

and across 2ohm, V = 12*2 = 24V

Now going back to fig 4

We can see that same current 12A is flowing through the circuit part 6ohm and 3 ohm.

Now when resistance are in parallel then voltage across them is same, so 6ohm and 3ohm resistance have same voltage, we can calculate the current from them

So, current across 6ohm resistance, I = V/R = 24/6= 4A

And current across 3ohm, I = 24/3 = 8A

Going to fig 3

As 2ohm and 4ohm resistance are in series then

Same current flows through them and voltage drop will be there

So, the voltage here is 24V and current is 4A

To find the individual voltage across 2ohm and 4ohm resistance

Voltage across 2ohm resistance, V =IR = 4*2 = 8V

And voltage across 4ohm,V = 4*4 = 16V

Similarly,Going to fig 2, voltage across 3ohm and 6ohm is 8V.

Current across 3ohm and 6ohm resistance are 8/3A and 8/6A

Going to fig 1

Current across 2ohm resistance is 8/3A