- The voltages dropped across three resistors (R
_{1}, R_{2}and R_{3}) connected in series are 5 V, 7 V and 10 V respectively. The supply current is 2 A. Calculate the supply voltage and the values of the three resistors.

*V*_{SUPPLY}= 5 V + 7 V + 10 V =**22 V**

*R*_{1}=5 V = **2.5 Ω**2 A

*R*_{2}=7 V = **3.5 Ω**2 A

*R*_{3}=10 V = **5 Ω**2 A

- For the circuit shown below, given that the voltage dropped across R
_{2}is 5 V and that the voltage dropped across R_{3}is 3 V, determine the value of the voltage dropped across R_{1}. If the total circuit resistance is 36 Ω, calculate the supply current and the resistor values.

*V*_{R1}= 18 V - (5 V + 3 V) =**10 V**

*I*_{SUPPLY}=18 V = **0.5 A**36 Ω

*R*_{1}=10 V = **20 Ω**0.5 A

*R*_{2}=5 V = **10 Ω**0.5 A

*R*_{3}=3 V = **6 Ω**0.5 A

- Two resistors are connected in series across an 18 V supply with a current of 5 A flowing through the circuit. If one of the resistors (R
_{1}) has a value of 2.4 Ω, determine the value of the other resistor (R_{2}), and the voltage dropped across R_{1}.

*R*_{TOTAL}=18 V = **3.6 Ω**5 A *R*_{2}= 3.6 Ω - 2.4 Ω =**1.2 Ω***V*_{R1}= 2.4 Ω × 5 A =**12 V**

- Imagine that you have a 100 Ω resistor (R
_{1}). You want to add a resistor (R_{2}) in series with R_{1}in order to limit the current to 0.5 amps when 110 volts is placed across the two resistors in series. How much resistance should you use?

*R*_{TOTAL}=110 V = **220 Ω**0.5 A *R*_{2}= 220 Ω - 120 Ω =**120 Ω**

- Two resistors (R
_{1}and R_{2}) of value 4 Ω and 12 Ω are connected in parallel across a 9 V battery. Calculate the resistance of the parallel branch, the supply current, and the current flowing through each resistor.

*R*_{TOTAL}=4 Ω × 12 Ω = **3 Ω**4 Ω + 12 Ω

*I*_{SUPPLY}=9 V = **3 A**3 Ω

*I*_{R1}=9 V = **2.5 A**4 Ω

*I*_{R2}=9 V = **0.75 A**12 Ω

- Find the total resistance between terminals A and B of the circuit of the circuit shown below.

*R*_{PARALLEL}=6 Ω × 18 Ω = **4.5 Ω**6 Ω + 18 Ω *R*_{TOTAL}= 2 Ω + 4.5 Ω + 1.5 Ω =**8 Ω**

- What is the equivalent resistance for this resistor combination?

*R*_{EQUIVALENT}=20 Ω × 14 Ω = **8.235 Ω**20 Ω + 14 Ω

- Resistors of value 20 Ω, 20 Ω and 30 Ω are connected in parallel. Determine the value of the series resistance that must be added to the parallel combination to obtain a total resistance of 10 Ω. If the completed circuit dissipates a power of 0.36 kW, find the current flowing.

*R*_{PARALLEL}=1 = **7.5 Ω**1 + 1 + 1 20 Ω 20 Ω 30 Ω *R*_{SERIES}= 10 Ω - 7.5 Ω =**2.5 Ω**(Additional series resistance)*I*= √(360 W ÷ 10 Ω) =**6 A**(*Power*=*I*^{ 2}×*R*)