**Combination
Circuits (430 only)**

** **

Combination circuits combine the features of parallel circuits with those of series circuits. The key to surviving these is to keep in mind the distinctive features of those circuits in mind.

- What is the overall resistance of this circuit?

(1) First we tackle the parallel
part. The 500 and 700 W resistors are in parallel. Their equivalent resistance is R_{p}
= [500^{-1} + 700^{-1}]^{-1} = 292 W.

(2) But this parallel branch is ** in
series** with the 400 W
resistor. So the total resistance is R

- What
is the potential difference measured from
*a*to*b*?

A common
mistake is to assume it’s 12 V. But the 400 W resistor is in series with
the parallel branch of the circuit, and voltage is *not* constant in series.
But V_{400 }_{W}
= I R , so we just need to find the overall
current.

V_{T}
= I_{T} (R_{T})

12 = I_{T}
(692)

I_{T }=
0.0173 A or 17.3 mA.

V_{400
}_{W} = I R = 0.0173(400) = 6.94
V.

- What
currents would be measured within the parallel branch? (see diagram for I
_{1}and I_{2})

Voltage
is constant in a parallel branch. To find this voltage, use the result in (b).

V_{parallel}
= V_{T}_{ }- V_{400 }_{W}
= 12 - 6.94 = 5.06 V.

Now apply
Ohm’s Law:

I_{1}
= V/R_{1} = 5.06/500 = 0.0101 A or 10.1 mA

I_{2}
= V/R_{2} = 5.06/700 = 0.00723 A or 7.2 mA

(Note how
I_{1 }+ I_{2} = I_{T.} This is a good way of verifying
your answers.)

a. Find the total current in the following:

First,
we’ll redraw the circuit to make sure we realized that the 5 and 8 W resistor
are in series, but they in turn are in parallel to the 10 W
resistor. The parallel branch is then series with both the 3 and 2 W
resistors.

In short
R_{T} = 2 + 3 + R_{P}

R_{PARALLEL} = [10^{-1} + (5 + 8)^{-1}]^{-1 }= 5.65
W

R_{T}
= 2 + 3 + 5.65 W =10.65 W

V_{T }=
I_{T}R_{T }

I_{T }=
V_{T }/ R_{T }= 12/10.65 = 1.1 A

- What is the voltage drop across the 8 W resistor?

V_{PARALLEL}
= I_{T} R_{PARALLEL} = 1.1(5.65) =
6.22 V; voltage is constant in parallel, but the problem is that the 8 W resistor is in series with the 5 W resistor. So we have to divide this voltage
by the combined series resistance in that branch to get its current:

6.22/(5 + 8) = 0.478 A

V_{8}_{W} = I R
= 0.478(8) = 3.83 V