Question

# For the circuit shown in the figure, find the current passing through 10 Ω resistor.

A
53 A
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B
512 A
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C
56 A
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D
518 A
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Solution

## The correct option is D 518 ALet the potential of upper wire be x and the potential of lower wire be zero. From the figure, we can write that i1+i2+i3+i4+5=10 x1+x2+x10+x5+5=10 ⇒x=5018 volts ∴i10=xR=(5018)110=518 A Hence, option (a) is the correct answer. Key concepts : Kirchhoff's Current Law (Junction law) This law is based on law of conservation charge. It states that "The algebraic sum of the currents meeting at a point of the circuit is zero' or total currents entering a junction equals to the total current leaving the junction. ∑Iin=∑Iout It is also known as KCL (Kirchoff's current law).

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