CK-12-Physics - Intermediate

http://www.ck12.org Chapter 17. Circuits

b. What is the battery current in the circuit?

Answer:

V=IR→ 20 V=I( 18 Ω)→Itotal= 1820 VΩ=^109 A

c. What is the current through each resistor?

Answer:

The 10Ωresistor:

The current has not split up until after passing the 10Ωresistor, therefore:

Itotal=^109 A

The 12Ωresistor:

We need to determine the voltage drop across the 12Ωresistor. This is done by determining the voltage drop across

the 10Ωresistor.

V 10 =IR=

( 10

9 A

)

( 10 Ω) =^1009 V

The voltage drop across the 12Ωresistor must beV 12 = 20 V−^1009 V=^809 V.

The current through the 12Ωresistor is, therefore:

V=IR→^809 V=I 12 ( 12 Ω)→I 12 =^2027 A

The 24Ωresistor:

We will solve this two ways:

1. Since we know the total currentItotalin the circuit, and we know the current through the 12ΩresistorI 12 :
   Itotal=I 12 +I 24 →I 24 =^109 A−^2027 A=^1027 A


2. The voltage drop across the 24Ωresistor,

( 80

9 V

)

, is the same as the voltage drop across the 12Ωresistor since

they are in parallel, therefore:

V=IR→^809 V=I 24 ( 24 Ω)→I 24 =^1027 A

Notice that the current in the 24Ωresistor is half of that in the 12Ωresistor. Since the resistors are in parallel, the

voltage drop across each resistor is the same, thus:

I 12 R 12 =I 24 R 24 →II^1224 =RR^2412

Therefore, a resistor with twice the resistance will have half the current.

If a very large resistance is in parallel with a very small resistance, most of the current will pass through the small