LearnTronics

DC Circuit Review and Practice

Tutorial Index

Review what we have built so far

No new theory on this page. This is a chance to practice the relationships until they begin to feel familiar.

Mental references worth carrying around:
1 A = 1 C/s
1 V = 1 J/C
V = IR
P = VI
1 hp = 746 W ≈ 0.75 kW

Practice test

These are intentionally similar to short state-test questions. Most should require no calculator.

One ampere is equal to: 1 joule per coulomb 1 coulomb per second 1 watt per volt 1 ohm per second

1 A = 1 C/s.

A 12 V source is across 4 Ω. Current is: 2 A 3 A 4 A 48 A

I = V/R = 12/4 = 3 A.

Two series resistors are 2 Ω and 6 Ω. Total resistance is: 3 Ω 4 Ω 8 Ω 12 Ω

Series resistances add: 2 + 6 = 8 Ω.

Two equal 8 Ω resistors are in parallel. Their equivalent resistance is: 4 Ω 8 Ω 16 Ω 64 Ω

Two equal resistors in parallel have half the resistance: 8/2 = 4 Ω.

A 10 V load draws 2 A. Power is: 5 W 10 W 12 W 20 W

P = VI = 10 × 2 = 20 W.

About how many kilowatts is 4 hp? 1 kW 2 kW 3 kW 4 kW

Using the mental shortcut 1 hp ≈ 0.75 kW, 4 hp ≈ 3 kW.

8 A enters a node. Branch currents are 3 A and ? What is the missing current? 3 A 5 A 8 A 11 A

KCL: 8 = 3 + 5.

A 20 V source has drops of 6 V and 4 V plus one unknown drop. The unknown is: 8 V 10 V 12 V 14 V

KVL: 20 - 6 - 4 = 10 V.

When a question takes two steps

Do not hunt for one giant formula. Write down what you know, solve the first easy relationship, then use that answer in the next relationship.

Example: if you know voltage and total resistance, first find total current. Once current is known, use it to find a resistor's voltage drop or power.

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