2.3 Circuit elements 17
resistors in series is therefore the sum of the three individual resistances. In
general, for n resistors is series, R~q = R~ + R2 + "" + Rn. In short this can be
written
n
Req = a~,=lRa (2.6)
Example 2.6
Determine (1) the current flowing in the circuit of Fig. 2.7, (2) the voltage
across each resistor.
Figure 2.7
I 5~ 10~ 20~ 15~
I! I I! I I I
R1 R2 R3 R4
200V
Solution
1 Using Ohm's law I = V/Req and from Equation (2.6),
Req = R 1 + R 2 + R 3 + R4, so Req = 5 + l0 -+- 20 + 15 = 50 11. Therefore
I = 200/50 = 4 A
2 Again, from Ohm's law VR1 = IR 1 = 4 • 5 = 20 V
VR2 -- IR 2 -- 4 x 10 = 40 V
VR3 = IR3 = 4 • 20 = 80V
VR4 = IR4 = 4 X 15 = 60V
Note that these add up to 200 V, which is the voltage of the supply.
Voltage division
I i al
i l<V1
R2
F----t
V2
Figure 2.8
For the two resistors shown connected in series in Fig. 2.8, V - I[R 1 + R2]. Also
V 1 -= Ig 1 so that