In the circuit shown in Fig `7.25`,
(a) what must the `emf epsilon` of the battery be in order for a current of `2.00 A` to flow through the `5.00 V` battery , as shown ? In the polarity of the battery correct as shown ?
b. How long does it take for `60.0 J` of thermal energy to be produced in the `10.0 Omega `resistor?

Answer

(a) `108.75 V` Polarity should be reversed
(b) `13.5 s`

Explanation

(a) First , do series // parallel reduction . Now, apply kirchhoff's laws and solve for `e`.
`Delta V _( adefa) = 0 or - ( 20 Omega ) ( 2 A) - 5 V - ( 20 Omega) I_(2) = 0 or I_(2) = -2.25 A`
` I_(1) + I_(2) = 2 A or I_(1) = 2 A -(-2.25 A) = 4.25 A`
Delta V _(adefa) = 0 or ( 15 Omega) ( 4.25 A) + epsilon - ( 20 Omega) ( - 2.25 A) = 0`
` epsilon = - 108.75 V , Polarity should be reversed.
(b) Parallel branch has a `10 Omega` resistance.
`Delta V_( Par) = RI = ( 10 Omega) ( 2 A) = 20 V`
Current in upper part is `I = ( Delta V)/( R) = ( 20 V)/( 30 Omega) = (2)/( 3) A`
`Pt = U or I^(2) Rt = U`
or ((2)/(3) A)^(2) ( 10 Omega )t = 60 J or t = 13.5 s`