-
The Norton equivalent across BB′ is given by—
-
-
RN = 50 Ω, Isc = 2 amp 26 5 -
RN = 50 Ω, Isc = 1 amp 13 5 -
RN = 50 Ω, Isc = 2 amp 13 5 - None of the above
-
Correct Option: B
Calculation for RN (i.e. Rth):
The equivalent circuit is given by 
RN = (20 || 10 + 10) || 5
| = |  | 20 | + 10 |  | || 5 = 50 3 || 5 |
| 3 |
| = | = | = | Ω | (50 / 3) + 5 | 65 | 13 |
Calculation for Isc
The equivalent circuit drawn in order to calculate I sc first calculate I.
| I = | Req |
where R veq = the equivalent resistance as seen from the terminal AA′
or R eq = 20 + 10 || 10 = 20 + 5 = 25 Ω
| I = | = | amp. (Given, V = 10 Volt) | 25 | 5 |
| = | amp. | 5 |