Question

please anyone answer me please anyone answer my question
this is a 3 mark question ​

Answer 1

Given :-

Four resistors of 16 Ω each are connected in parallel, and another four resistors of 16 Ω each are connected in series.

To determine :-

The total resistance.

Acknowledgement :-

For connection in series,

$\rm{R_{Equivalent}=R_1+R_2+R_3+....+R_n}$

For connection in parallel,

$\displaystyle{\rm{\frac{1}{R_{Equivalent}}=\frac{1}{R_1} +\frac{1}{R_2} +\frac{1}{R_3}+.....+\frac{1}{R_n} }}$

Solution :-

Applying the formulas above :-

For parallel connection :-

$\displaystyle{\rm{\frac{1}{R_{Equivalent}}=\frac{1}{R_1} +\frac{1}{R_2} +\frac{1}{R_3}+\frac{1}{R_4} }}$

$\displaystyle{\rm{\longrightarrow \frac{1}{R_{Equivalent}}=\frac{1}{16} +\frac{1}{16} +\frac{1}{16}+\frac{1}{16} }}$

$\displaystyle{\rm{\longrightarrow \frac{1}{R_{Equivalent}}= \frac{4}{16} }}\\\\\\\displaystyle{\rm{\longrightarrow \frac{1}{R_{Equivalent}}= \frac{1}{4} }}\\\\\\\displaystyle{\rm{\longrightarrow R_{Equivalent}= 4\ \Omega }}$

According to question, 4 such parallel connections are again connected in series.

Therefore, the total resistance = 4 + 4 + 4 + 4

⇒Total resistance = 16 Ω

Therefore, the total resistance is 16 Ω.

Answer 2

Answer:

16 ohms

Explanation:

Given:

• Four resistances of 16 ohms are connected in parallel
• Four combinations such as the above are connected in series

To find:

• Equivalent resistance across the circuit

First we need to find the equivalent resistance for four resisitors in parallel and then connect such four in series

Parallel combination:

Equivalent resistance when four 16 ohms resistors are connected in parallel:

$\frac{1}{Req} =\frac{1}{16} +\frac{1}{16} +\frac{1}{16} +\frac{1}{16}$

$\frac{1}{Req}=\frac{4}{16}$

$Req=\frac{16}{4}$

$Req=4 \ ohms$

Four combinations of 4 ohms are connected in series, so:

Series combination:

$Req=4+4+4+4$

$Req=16 \ ohms$

The total resistance is equal to 16 ohms