Question

please do this.....​

Answer 1

Answer:

6.0083..Ω

Explanation:

Given:

• 1.6Ω resistor in series with the other two resistors
• 12.4Ω and 6.84Ω in parallel and 1.6Ω in series to them both

To find:

• Equivalent or net resistance across the circuit

When any two resistors are in parallel:

$Req = \frac{R1 \times R2}{R1+R2}$

Here:

Req = $\frac{12.4 \times 6.84}{12.4+6.84}$

Req =$\frac{84.816}{19.24}$

Req=4.4083... Ω

Now this is in series with 1.6Ω, so:

Req= 4.4083 +1.6

Req = 6.0083..Ω (approx)

The approximate equivalent resistance is equal to 6.0083...Ω

Answer 2

Given :

$R_1=1.6\Omega \:\:\:, \:\:\:R_2=6.84\Omega\:\:\:,\:\:\:R_3=12.4\Omega$

To Find :

$R_{resultant}=?$

Formula :

When resistances are connected in series and parallel .The formulas are as follows :

$R_{parallel}=\frac{R_1.R_2}{R_1+R_2} \\\\R_{series}=R_1+R_2$

Explanation :

$R_2 \:\:\:and\:\:\:R_3\:\:\:are\:\:in\:\:parallel.\\\\=>R_{23}=\frac{R_2.R_3}{R_2+R_3}\\\\ =>R_{23}=\frac{6.84*12.4}{6.84+12.4} \\\\=>R_{23}=4.4\:\Omega$

Now $R_{23}\:\:and\:\:R_1$ are in series.

$=>R_{123}=R_1+R_{23}\\\\=>R_{123}=1.6+4.4\\\\=>R_{123}=6\:\Omega\\\\=>R_{resultant}=6\:\Omega$