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

Explanation

(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`