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in a circuit there is a battery of 10v and a resistor of 5ohm which is connected in series with two resistors of 10 ohm each that are connected parallely. Find 'I' and 'V' across each resistor.

Let a = 5 ohmb = 10 ohmc = 10 ohmwhen b and c are connected in parallel. their equivalent resistance is d:d = 10×1010+10  = 5 ohmNow A and d are connected in series. Net resistance of the circuit is :R = a + d = 5+5 = 10 ohmV = 10 vcurrent in the circuitV = IRI = V/R = 10/10 = 1 AAs in series current remains the same, current through the 5 ohm is 1 A and thorugh d is 1A.Voltage across 5 ohm is:V = 1×5 = 5 vVoltage across d is:V = 1×5 = 5 voltSince b and c are connected in parallel, volatge acroos them will remains same.Volatge across b is 5 voltCurrent across B is Ib = V/b = 5/10 = 0.5 Asimilarly volatge across c isIc = V/c = 5/10 = 0.5 A

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first calculate net resistance i.e. 50 ohm+1/(1/10+1/10)

Rnet=55 ohm

then use V=IR

10=55I

Inet=0.19 A (approx.)

in series current remains constant and V diifers

thus resister of 50 ohm will recieve 0.19A current.

using V=IR, V=50*0.19=9.5V

therefore the resisters in parallel will recieve 10-9.5=0.5V

in parallel, current differs but V remains constant

thus both 10 ohm resisters will recieve 0.5V voltage

using I=V/R in 1 10 ohm resistor

=I=0.5/10=0.05A current

since both resistors in parallel have same resistance, both will have 0.5V and 0.05A

THEREFORE in R1, I=0.19A,V=9.5V

in R2, I=0.05A, v=0.5V

in R3, I=0.05A, v=0.5V

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