Question

a) Two 8kΩ resistors are connected in parallel. Calculate the equivalent
resistance,
b) Two 8kΩ resistors are connected in series. Calculate the equivalent resistance.


❌❌ Don't spam ❌❌
​

Answer 1

Answer:

• Equivalent Resistance in Parallel is 4 KΩ
• Equivalent Resistance in Series 16 KΩ

Explanation:

Given:-

• Two Resistors of 8 KΩ ( In Parallel)
• Two Resistors of 8 KΩ ( In Series)

To Find:-

• Equivalent Resistance in Parallel ,Rs
• Equivalent Resistance in Series ,Rp

Solution:-

Firstly we calculate the Equivalent resistance in Parallel.

• 1/Rp = 1/R1 + 1/R2

Where,

• Rp denote equivalent Resistance in Parallel
• R1 denote equivalent resistance of Resistor
• R2 denote equivalent resistance of Resistor

Substitute the value we get

→ 1/Rp = 1/8 + 1/8

→ 1/Rp = 2/8

→ 1/Rp = 1/4

→ Rp = 4 KΩ

• Hence , equivalent resistance in Parallel is 4 KΩ

Now, Calculating the Equivalent resistance in Series.

• Rs = R1 + R2

where

• Rs denote equivalent resistance in Series
• R1 denote resistance of Resistor
• R2 denote resistance of Resistor

Substitute the value we get

→ Rs = 8 KΩ+ 8 KΩ

→ Rs = 16 KΩ

• Hence, Equivalent resistance in Series is 16 KΩ.

Answer 2

Answer:

Given :-

• Two resistor of resistance = 8kΩ (Parallel)
• Two resistor of resistance = 8kΩ (Series)

To Find :-

In both equivalent resistance

Solution :-

For parallel series we will use

$\huge \bf \: \frac{1}{Rp} = \frac{1}{R1} + \frac{1}{R2}$

$\tt \dfrac{1}{Rp} \: = \dfrac{1}{8} + \dfrac{1}{8}$

$\tt \dfrac{1}{Rp} \: = \dfrac{2}{8}$

$\tt \dfrac{1}{Rp} \: = \dfrac{1}{4}$

$\tt \: Rp \: = 4 \:$

Equivalent resistance in parallel = 4kΩ

For series

$\tt \: Rs \: = R1 + R2$

$\tt \: Rs \: = 8 + 8$

$\tt \: Rs \: = 16$

Equivalent resistance in series = 16kΩ