In the circuit shown in the circuit thro...

In the circuit shown in the circuit through

the `3 Omega` resistor is `0.50A`

the `3 Omega` resistor is `0.25A`

the `4 Omega` resistor is `0.50A`

the `4 Omega` resistor is `0.25A`

Text Solution

Verified by Experts

The correct Answer is:

The current through the various branches of the circuit willl be

According to Kirchhoff's second law in closed circuit `CDEFC`
`2I_(2)+4I_(2)+ 2I_(2) - 8(I_(1)- I_(2)) = 0`
or `8 I_(1) = 16I_(2) or I_(1)= 2I_(2)`....(i)
In closed circuit `BCRFGB` ,
`2I_(2)+8(I_(1)- I_(2)) +2I_(2) - 8(I- I_(1)) = 0`
or `20 I_(1) - 8 I_(2)= 8I or 40 I_(2)- 8I_(2) = 8I`
or `l_(2)= l//4`....(ii)
From (i) `I_(1) = 2 xx I//4= I//2`
In a closed circuit `ABGHA`
`(3 I+8(I - I_(1))+2 I= 9`
or `9 = 13 I - 8 I_(1)= 13 I - 8(I//2) = 9 I or I = 1 A`
From (ii) `I_(2) = (1)/(4) = 0.25A`

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For the circuit shown

Current in circuit in 10A

Voltage across inductor is 100V

Voltage acros capacitor is less than that of supply voltage

Voltage across capacitor is more than that of supply voltage

In the circuit shown in the figure .

current passing through `2Omega` resistance is zero.

current passing through `4Omega` resistance is `5A`

current passing through `5Omega` resistance is 4A.

potential difference across `2Omega` resistance is zero.

In the circuit shown in the figure,

The current through NP is `0.5` A

The value of `R_(1)=40Omega`

The value of R =14`Omega`

sThe potential difference across R = 49 V