NEET UG 2006 — Physics Electromagnetism

$1 \Omega$1Ω and $3 \Omega$3Ω are in series so, $$\begin{aligned} & R_s=R_1+R_2 \\ & R_s=3+1=4 \Omega \end{aligned}$$​Rs​=R1​+R2​Rs​=3+1=4Ω​ and $R_2=8 \Omega$R2​=8Ω net $i=$i= current in the circuit Current through $R_1$R1​ is $i_1=\frac{i \times R_2}{R_1+R_2}$i1​=R1​+R2​i×R2​​ $=\frac{i \times 8}{12}=\frac{21}{3}$=12i×8​=321​ Current $R_2$R2​ is $i_2=\frac{1 \times R_1}{R_1+R_2}=\frac{i \times 4}{12}$i2​=R1​+R2​1×R1​​=12i×4​ $=\frac{i \times 8}{12}$=12i×8​ Power dissipated in $3 \Omega$3Ω resistance is: $P_1=l_1^2 \times 3$P1​=l12​×3 Power dissipated in $8 \Omega$8Ω resistance is: $P_2=l_2^2 \times 8$P2​=l22​×8 For (i) and (ii) $$\begin{aligned} & \frac{P_1}{P_2}=\frac{i_1^2 \times 3}{i_2^2 \times 8} \\ & \Rightarrow \frac{P_1}{P_2} \end{aligned}$$​P2​P1​​=i22​×8i12​×3​⇒P2​P1​​​ $$\begin{aligned} & =\frac{\left(\frac{2 i}{3}\right)^2 \times 3}{\left(\frac{i}{3}\right)^2 \times 8}=\frac{3}{2} \\ \therefore \quad P_1 & =\frac{3}{2} \times P_2=\frac{3}{2} \times 2 =3 \mathrm{~W} \end{aligned}$$∴P1​​=(3i​)2×8(32i​)2×3​=23​=23​×P2​=23​×2=3 W​ $$\begin{aligned} \because & P & =100 \mathrm{~W}, V=125 V \\ \therefore & P & =V I \\ \Rightarrow & I & =\frac{P}{V}=\frac{100}{125} A \end{aligned}$$∵∴⇒​PPI​=100 W,V=125V=VI=VP​=125100​A​