Question

What is the highest and lowest total resistance that can be secured by the combinations of 2 ohm,4 ohm,6 ohm and 8 ohm. ​

Answer 1

We have to find the highest and lowest total resistance that can be secured by the combinations of 2 ohm,4 ohm,6 ohm and 8 ohm.

Take R1 = 2ohm, R2 = 4ohm, R3 = 6ohm and R4 = 8ohm

There are two ways to find the resistance. First by using series combination and second by using parallel combination.

To find lowest resistance we have to add the resistors in the parallel combination and to find the highest resistance we have to add the resistors in the series combination.

For highest resistance:

Rs = R1 + R2 + R3 + R4

Rs = 2 + 4 + 6 + 8

Rs = 2(1 + 2 + 3 + 4

Rs = 2(10)

Rs = 20 ohm

For lowest resistance:

1/Rp = 1/R1 + 1/R2 + 1/R3 + 1/R4

1/Rp = 1/2 + 1/4 + 1/6 + 1/8

1/Rp = (12 + 6 + 4 + 3)/24

1/Rp = 25/24

Rp = 24/25

Rp = 0.96 ohm

Therefore, total highest resistance is 20 ohm and total lowest resistance is 0.96 ohm.

Answer 2

$\maltese$ Answer $\maltese$

$\checkmark$ Given:

$\star$ Four Resistances

• R₁ = 2 Ω
• R₂ = 4 Ω
• R₃ = 6 Ω
• R₄ = 8 Ω

$\checkmark$ To Find:

The combinations of the given Resistances to get,

• Highest Resistance
• Lowest Resistance

$\checkmark$ Solution:

We know that,

In Combination of Resistances,

There are two types of combinations,

• Series Combination
• Parallel Combination

Basically,

Series Combination of Resistances

In Series Combination of Resistances

• Resistances are connected along a same line
• Value of Current remains same

The Equivalent Resistance in Series combination is given by the formula

$\orange{\boxed{\boxed{\sf R_{s}=R_{1}+R_{2}+R_{3}+R_{4}}}}$

Parallel Combination of Resistances

In Parallel Combination of Resistances

• Resistances are connected across each other
• Value of Voltage remains same

The Equivalent Resistance in Parallel combination is given by the formula

$\red{\boxed{\boxed{\sf \dfrac{1}{R_{p}}=\dfrac{1}{R_{1}}+\dfrac{1}{R_{2}}+\dfrac{1}{R_{3}}+\dfrac{1}{R_{4}}}}}$

NOTE:

For any given Resistances,

Highest Resistance will be in Series Combination

Lowest Resistance will be in Parallel Combination

Therefore,

Highest Resistance will be observed in the Series Combination

Given that,

Four Resistances,

• R₁ = 2 Ω
• R₂ = 4 Ω
• R₃ = 6 Ω
• R₄ = 8 Ω

Hence,

Using the formula given above,

$\blue{\implies \sf R_{s}=R_{1}+R_{2}+R_{3}+R_{4}}$

Substituting the above values in the Formula,

We get,

$\orange{\implies \sf R_{s}=2 \; \Omega +4 \; \Omega +6 \; \Omega+8 \; \Omega }$

$\orange{\implies \sf R_{s}=20 \; \Omega }$

Therefore,

$\star$ Highest Resistance = 20 Ω

Similarly,

Lowest Resistance = Parallel Combination

Therefore,

Lowest Resistance will be observed in the Parallel Combination

Given that,

Four Resistances,

• R₁ = 2 Ω
• R₂ = 4 Ω
• R₃ = 6 Ω
• R₄ = 8 Ω

Hence,

Using the formula given above,

$\pink{\implies \sf \dfrac{1}{R_{p}}=\dfrac{1}{R_{1}}+\dfrac{1}{R_{2}}+\dfrac{1}{R_{3}}+\dfrac{1}{R_{4}}}$

Substituting the above values in the Formula,

We get,

$\red{\implies \sf \dfrac{1}{R_{p}}=\dfrac{1}{2 \Omega}+\dfrac{1}{4 \Omega}+\dfrac{1}{6 \Omega}+\dfrac{1}{8 \Omega}}$

$\red{\implies \sf \dfrac{1}{R_{p}}=\dfrac{12+6+4+3}{24} \ \Omega}$

$\red{\implies \sf \dfrac{1}{R_{p}}=\dfrac{25}{24} \ \Omega}$

$\red{\implies \sf \dfrac{R_{p}}{1}=\dfrac{24}{25} \ \Omega}$

$\red{\implies \sf R_{p}=\dfrac{24}{25} \ \Omega}$

$\red{\implies \sf R_{p}=0.96 \ \Omega}$

Therefore,

$\star$ Lowest Resistance = 0.96 Ω

So, Finally,

$\checkmark$ Highest Resistance = 20 Ω

$\checkmark$ Lowest Resistance = 0.96 Ω