Question

ANSWER THE QUESTION IN ATTACHMENT​

Answer 1

Given

• Here we have seven 7Ω resistors and one 2Ω resistor
• They are connected in a specific way

To Find

• Equivalent Resistance between A & B

Solution

● 1/Rₙₑₜ = 1/R₁ + 1/R₂ .... 1/Rₙ [Parallel Connection]

● Rₙₑₜ = R₁ + R₂ .... Rₙ [Series Connection]

● All the seven 7Ω resistors are connected in parallel, they all together are connected in series with the 2Ω resistor

✭ Parallel Connection :

• Equivalent Resistance of seven, 7Ω resistors in parallel

→ 1/Rₙₑₜ = 1/R₁ + 1/R₂ .... 1/Rₙ

→ 1/Rₙₑₜ = 1/7 + 1/7 + 1/7 + 1/7 + 1/7 + 1/7 + 1/7

→ 1/Rₙₑₜ = (1+1+1+1+1+1+1)/7

→ 1/Rₙₑₜ = 7/7

→ Rₙₑₜ = 7/7

→ Rₙₑₜ = 1Ω

━━━━━━━━━━━━━━━━━━

✭ Equivalent Resistance :

• So now that we have the equivalent Resistance of the Parallel Connection thry are connected in series with the 2Ω resistor

→ Rₙₑₜ = R₁ + R₂ .... Rₙ

→ Rₙₑₜ = 2 + 1

→ Rₙₑₜ = 3Ω

Answer 2

Given:-

✹ Seven - 7 ohm resistors

✹ One - 2 ohm resistors

To Prove:-

✹ Equivalent resistors between A and B

Solution:-

✹ To Find Parallel resistors:

✹ Using Formula,

$\large \star{ \boxed{ \boxed{ \frak{ \purple{ \frac{1}{R_n \:_e \:_t } = \frac{1}{R_1 } + \frac{1}{R_2}.... \frac{1}{R_n} }}}}}$

✹ Substitute the value in the formula,

$\star{ \boxed{ \boxed{ \frak{ \frac{1}{R_n \:_e \:_t } = \frac{1}{7 } + \frac{1}{7} + \frac{1}{7} + \frac{1}{7} + \frac{1}{7} +\frac{1}{7} +\frac{1}{7} \: \: }}}}$

$\star{ \boxed{ \boxed{ \frak{ \frac{1}{R_n \:_e \:_t } = \frac{1 + 1 + 1 + 1 + 1 + 1 + 1}{7 } }}}}$

$\star{ \boxed{ \boxed{ \frak{ {R_n \:_e \:_t } = 1 \: ohm }}}}$

✹ To Find Equivalent resistors:

✹ Using Formula,

$\large\star{ \boxed{ \boxed{ \frak{ \purple{ {R_n \:_e \:_t } = R_1 + R_2.... R_n}}}}}$

✹ Substitute the value in the Formula,

$\star{ \boxed{ \boxed{ \frak{ {R_n \:_e \:_t } = 2 + 1 = 3}}}}$

$\star{ \boxed{ \boxed{ \frak{ {R_n \:_e \:_t } = 3 \: ohm}}}}$

✹ Parallel resistors = 1 ohm

✹ Equivalent resistors = 3 ohm