For the circuit shown in figure, the Thevenin voltage across the terminals a and b is

20 V

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30 V

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-10 V

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15 V

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Solution

The correct option is B 30 V

Apply nodal analysis,
$\frac{V_{1} - 32}{4} + \frac{V_{1}}{12} - 2 = 0$
$V_{1} (\frac{1}{4} + \frac{1}{12}) = 10$
$V_{1} (\frac{4}{12}) = 10$
$V_{1} = \frac{120}{4} = 30 V$
The thevenin's voltage across a and b, as it is open circuited, will be
$V_{t h} = V_{1} = 30 V$

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For the circuit shown in figure, the Thevenin voltage across the terminals a and b is

The correct option is B 30 V Apply nodal analysis, V1−324+V112−2=0 V1(14+112)=10 V1(412)=10 V1=1204=30V The thevenin's voltage across a and b, as it is open circuited, will be Vth=V1=30V

The correct option is B 30 V

Apply nodal analysis,
$\frac{V_{1} - 32}{4} + \frac{V_{1}}{12} - 2 = 0$
$V_{1} (\frac{1}{4} + \frac{1}{12}) = 10$
$V_{1} (\frac{4}{12}) = 10$
$V_{1} = \frac{120}{4} = 30 V$
The thevenin's voltage across a and b, as it is open circuited, will be
$V_{t h} = V_{1} = 30 V$