Circuits homework solutions

The equivalent  0  , a short circuit. 07 resistance is The voltage sources are connected in series and can be replaced by a single equivalent voltage source. After doing so, and labeling the resistor currents, we have the circuit shown. fThe parallel current sources can be replaced by an equivalent current source.

Solution: i or R1  8 8 R1 i 8  R 2 (2  i )  i  2  or R 2  8 8 R2 2i a i  2  A ; R1   12  8 2 8 6 3 2  b  i   A ; R 2  4  12  3 8 4 8 6 3 2  c  R1  R 2 will cause i= 2  1 A.

(a) Determine the value of the resistance R in Figure P 3.3-2b that makes the circuit in Figure P 3.3-2b equivalent to the circuit in Figure P 3.3-2a. (b) Determine the current i in Figure P 3.3-2b. Because the circuits are equivalent, the current i in Figure P 3.3-2a is equal to the current i in Figure P 3.3-2b.

Solution: 48  24 a R  16   32  48  24 32  32 b v  32  32 24  16 V ; 32  32 8 32  32 i  A 16 1 32 2 c i2    A 48 1 1 48  24 2 3 P 3.6-4 (a) Determine values of R1 and R2 in Figure P 3.6-4b that make the circuit in Figure P 3.6-4b equivalent to the circuit in Figure P 3.6-4a.