Using Kirchhoff's rules in the circuit showndetermine (a) the voltage drop across the unknown resistor R, and (b) the current I in the arm EF.

Using Kirchhoff's rules in the circuit showndetermine (a) the voltage

1.	Using Kirchhoff's rules in the circuit showndetermine (a) the voltage drop across the unknown resistor R, and (b) the current I in the arm EF.
Answer» Solution :(a) As per Kirchhoff.s first law CURRENT flowing through unknown resistor R is (I + 1) A. As per Kirchhoff.s second law for MESH ACDBA, we have - (I + 1)R – 1.2 + 3 = 0 or (I + 1)R=1 ... (i) ` therefore` VOLTAGE drop across R =V=(1+1)R=1 Volt (b) Again for mesh ECDFE, we have - (I + 1)R-I.3 + 5 =0 or (I + 1)R =5 - 3I...(ii) Comparing (i) and (ii), we get `5 - 3I= 1 RARR I = 4/3 A = 1.33 A`

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Using Kirchhoff's rules in the circuit showndetermine (a) the voltage drop across the unknown resistor R, and (b) the current I in the arm EF.

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