Question

Two resistors R₁ and R₂ of resistance 3Ω and 6Ω respectively are connected in parallel across a battery of potential difference 12 V. Calculate the electrical energy consumed in 1 min. in each resistance.​

Answer 1

$\bf \to \: \green{{ \underline{given \div }}}$

$\sf \to \: two \: resistors \: r_{1} \: and \: r_{2} \: are \: connected \: in \: parallel \\ \\ \sf \to \: resistance \: = 3 \Omega \: and \: 6 \ Omega \\ \\ \sf \to \: potential \: difference \: = 12v$

$\bf \to \: \orange{{ \underline{to \: find \div }}}$

$\sf \to \: to \: find \: the \: electric \: energy \: consume \: in \: 1 \: min \: in \: each \: resistor.$

$\bf \to \: { \underline{solution \div }}$

$\sf \to \: for \: parallel \: combination \\ \\ \sf \to \: r_{eq} \: = \dfrac{ r_{1} r_{2} }{ r_{1} \: + \: r_{2} } \\ \\ \sf \to \: r_{eq} \: = \dfrac{3 \times 6}{3 + 6} \\ \\ \sf \to \: r_{eq} \: = \frac{ \cancel{18}}{ \cancel{9}} = 2 \Omega$

$\\ \\ \sf \to \: from \: ohms \: law \: \implies \: v \: = ir \\ \\ \sf \to \: 12 = 2i \\ \\ \sf \to \: i \: = 6a \\ \\ \sf \to \: for \: resistor \: r_{1} \: \\ \\ \sf \to \: r_{1} \: = 3 \Omega \: and \: i \: = 6a \\ \\ \sf \to \: v \: = 6 \times 3 = 18v$

$\\ \\ \sf \to \: for \: resistor \: r_{2} \: \\ \\ \sf \to \: r_{2} \: = 6 \Omega \: and \: i \: = 6a \\ \\ \sf \to \: v \: = 6 \times 6 = 36v \\ \\ \sf \to \: power \: = v \: \times i \: \\ \\ \sf \to \: power \: for \: resistor \: r_{1} \: \\ \\ \sf \to \: power \: = \: v_{1} \: \times i \implies \: 18 \times 6 = 108w$

$\\ \\ \sf \to \: electric \: energy \: consumed \: in \: 1 \: minutes \\ \\ \sf \to \: power \: \times \: time \\ \\ \sf \to \: 108 \times 60 = 6480j$

$\\ \\ \sf \to \: power \: for \: resistor \: r_{2} \: \\ \\ \sf \to \: v_{2} \: \times i \implies \: 36 \times 6 = 216w \\ \\ \sf \to \: energy \: consumed \: in \: 1 \: minutes \\ \\ \sf \to power \: \times time \\ \\ \sf \to \: 216 \times 60 = 12960 \: j$