For a two-branch current divider we have i1 R2 ¼ i R1 þ R2 This may be expressed as follows: The ratio of the current in one branch of a two-branch parallel circuit to the total current is equal to the ratio of the resistance of the other branch resistance to the sum of the two resistances. EXAMPLE 3.8. A current of 30.0 mA is to be divided into two branch currents of 20.0 mA and 10.0 mA by a network with an equivalent resistance equal to or greater than 10.0 . Obtain the branch resistances. 20 mA R2 ¼ 30 mA R1 þ R2
10 mA R1 ¼ 30 mA R1 þ R2
R1 R2 10:0
R1 þ R2
Solving these equations yields R1 15:0 and R2 30:0 .
Solved Problems 3.1
Find V3 and its polarity if the current I in the circuit of Fig. 3-7 is 0.40 A.
Fig. 3-7 Assume that V3 has the same polarity as V1 . Applying KVL and starting from the lower left corner, V1 Ið5:0Þ V2 Ið20:0Þ þ V3 ¼ 0 50:0 2:0 10:0 8:0 þ V3 ¼ 0 V3 ¼ 30:0 V Terminal b is positive with respect to terminal a.
Obtain the currents I1 and I2 for the network shown in Fig. 3-8. a and b comprise one node.
2:0 þ 7:0 þ I1 ¼ 3:0 Also, c and d comprise a single node.
I1 ¼ 6:0 A
4:0 þ 6:0 ¼ I2 þ 1:0
Find the current I for the circuit shown in Fig. 3-9.
I2 ¼ 9:0 A
Published on May 10, 2013