Question

plzz answer this question fast plz plz plz....correct answer plz otherwise will be reported​

Answer 1

★Solution:-

• First step is to connect the an imaginary battery across A and B
• Second step is to write the potential across the resistors
• Rearrange the circuit (as shown in the diagram )

From the diagram ,we can easily conclude that the two R 1 resistors are connected in parallel combination to each other and thier equivalent resistance is in series combination with R

As per the given data

• Equivalent resistance of the combination = 25 R /19
• R = 19

Two resistors R 1 and R 1 are connected in parallel with each other

Equivalent resistance for resistors connected in parallel combination is given by the formula

$\bigstar \boxed{\mathtt{\dfrac{1}{R_p} = \dfrac{1}{R_1}+ \dfrac{1}{R_2}+\dfrac{1}{R_3}.....+ \dfrac{1}{R_n}}}$

hence ,

$\implies\mathtt{ \dfrac{1}{R_p }= \dfrac{1}{R_1 }+ \dfrac{1}{R_1 }}$

$\implies\mathtt{ \dfrac{1}{R_p }= \dfrac{2}{R_1 }}$

$\implies\mathtt{ R_p = \dfrac{R_1}{ 2}}$

Now ,R p is in series with R

Equivalent resistance for resistors connected in series combination is given by the formula ,

$\bigstar\boxed{\mathtt{R_s = R_1 + R_2 +R_3 ....+R_n}}$

hence ,

$\implies\mathtt{R_{net }= R + R p }$

$\implies\mathtt{ \dfrac{25R}{19} = 19+ \dfrac{R_1}{2} }$

$\implies\mathtt{ \dfrac{25 \times \cancel{19}}{ \cancel{19}} = 19+ \dfrac{R_1}{2} }$

$\implies\mathtt{\dfrac{R_1}{2} = 25 -19}$

$\implies\mathtt{\dfrac{R_1}{2} = 6}$

$\implies \boxed{\mathtt{R_1= 12 \Omega}}$