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Q.

The potential difference between points A and B in the circuit shown in figure is 16V. then

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a

the current through the $2 Ω$ resistance is 3.5A

b

the potential difference between the terminals of they 9V battery is 7V

c

the current through the $4 Ω$ resistance is 2.5A

d

the current through the $3 Ω$ resistance is 1.5A

answer is A, C, D.

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Detailed Solution

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$V_{A} - 4 i_{1} - 9 - 1 (i_{1} - i_{2}) + 3 - 4 i_{1} = V_{B} 16 - 9 + 3 - 9 i_{1} + i_{2} = 0 10 - 9 i_{1} + i_{2} = 0 10 - 13.5 + i_{2} = 0 i_{2} = 3.5 A 16 - 4 i_{1} - 2 i_{2} + 3 - 4 i_{1} = 0 19 - 8 i_{1} - 2 i_{2} = 0 20 - 18 i_{1} + 2 i_{2} = 0 39 - 26 i_{1} = 0 i = \frac{39}{26} = 1.5 A$

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The potential difference between points A and B in the circuit shown in figure is 16V. then