## Resistances Problems and Solutions

- Category: High School
- Written by fisikastudycenter

The Examples of High School Physics Problems and Solutions : Resistances in Series-Parallel (Grade 10) and how to find total resistance of 12 identic resistors that forming a cube.

**Problem 1**

Given three resistors shown below,

Find the total resistance of A-B!

**Solution**

The three resistances are connected in series, so the total resistance is equal to the sum of the resistances of A-B:

R_{T} = 2 + 3 + 6 = 11 Ohm

**Problem 2**

Find the total resistance for three resistors below!

**Solution**

The three resistances are connected in parallel so using the parallel formula :

**Problem 3**

Ten resistors, each having a resistance of 10 Ω , are arranged in series-parallel combinations shown below.

Find the total resistance!

**Solution**

R_{23}, the sum of R_{2} and R_{3} (series) :

R_{46}, the sum of R_{4}, R_{5} and R_{6} (series) :

R_{710}, the sum of R_{7} , R_{8} , R_{9} and R_{10}(series) :

Using parallel formula for R_{1}, R_{23}, R_{46} and R_{710} give us R_{AB}:

1/R_{AB} = 1/10 + 1/20 + 1/30 + 1/40

1/R_{AB} = 12/120 + 6/120 + 4/120 + 3/120

Invert:

R_{AB} = 120 / 25 = 4,8 Ohm

**Problem 4**

Ten resistors, each having a resistance of 120 Ω, are arranged below.

Find the total resistance!

**Solution**

Using parallel formula for:

R_{2} and R_{3}. The result is R_{23} = 60 Ω

R_{4} , R_{5} and R_{6}. The result is R_{46} = 40 Ω

R_{7} , R_{8} , R_{9} and R_{10} . The result is R_{710} = 30 Ω

Series for R_{1} , R_{23} , R_{46} dan R_{710}. The result is R_{total} = R_{AB}

R_{AB} = 120 + 60 + 40 + 30 = 250 Ω

**Problem 5**

Find the total resistance of eight resistors below:

R_{1} = 10 Ω

R_{2} = 2 Ω

R_{3} = 3 Ω

R_{4} = 17 Ω

R_{5} = 20 Ω

R_{6} = 20 Ω

R_{7} = 8 Ω

R_{8} = 10 Ω

**Solution**

→ Series for R_{3} and R_{4} → R_{34}

R_{34} = R_{3} + R_{4} = 3 + 17 = 20 Ω

→ Parallel for R_{5} and R_{34} → R_{35}

R_{35} = 10 Ω

→ Series for R_{2}, R_{35} and R_{7} → R_{27}

R_{27} = 2 + 10 + 8 = 20 Ω

→ Parallel for R_{27} and R_{6} → R_{276}

R_{276} = 10 Ω

→ Series for R_{1} , R_{276} and R_{8} → R_{AB} or R_{total}

R_{AB} = 10 + 10 + 10 = 30 Ω

**Problem 6**

8 resistors, each having a resistance of 10 Ω, are arranged below!

Find the total resistance!

**Solution**

There will be no currents flow along R_{2} and R_{4} , so we could ignore them.

→ Series for R_{5} and R_{6} :

R_{56} = 20 Ω

→ Series for R_{7} and R_{8} :

R_{78} = 20 Ω

→ Parallel for R_{56} and R_{78} :

R_{58} = 10 Ω

→ Series for R_{1} , R_{58} and R_{3} to find R_{PQ} = R_{total} :

R_{PQ} = 10 + 10 + 10 = 30 Ω

and then, how to find total resistance of 12 identic resistors that forming a cube .

**Problem 7**

12 identic resistors are arranged and form a cube shown below.

Each resistor has 18 Ohm of resistance. Find the total resistance between P and Q!

**Solution**

Using the shortcut for this problem

R_{total} = ^{5}/_{6} R

R_{total} = ^{5}/_{6} (18) = 15 Ohm

**Try This!**

12 identic resistors are arranged and form a cube shown below.

Each resistor has 36 Ohm of resistance. Find the total resistance between P and Q!