Question

Pls solve this question

Answer 1

Answer:

(i) V₁ = ${144\over54}~V$

(ii)V₂ = ${504\over54}~V$

(iii) i₁ = ${336\over378}~A$

(iv) i₂ = ${252\over378}~A$

(v) i = ${84\over54}~A$

Explanation:

The equivalent resistance will come out to be

${4\times 3\over 4+3}+6={12\over 7}+6={54\over7}~\Omega$

Hence, According to Ohm's law, $V=IR$

$i={V\over R}={12\over {54\over7}}~Ampere={84\over54}~A$

Since, According to $V=IR$

$I$  ∝  $1\over R$

Hence, current distribution of 84/54 Amperes, in the parallel resistances, will be inversely proportional to their resistances. (In parallel, Voltage remains same an Current is different).

Hence,

$i_1={4\over 3+4}\times {84\over 54}={4\over 7}\times {84\over 54}={336\over378}~A$

And,

$i_2={3\over 3+4}\times {84\over 54}={3\over 7}\times {84\over 54}={252\over378}~A$

Now, the equivalent resistances of only the two parallel ones is

${4\times 3\over 4+3}={12\over 7}~\Omega$

So, this 12/7 Ω is in series with 6 Ω now. And we know that, in series, Current remains same and voltage is different.

Also, V ∝ R

So, voltage distribution of 12 V will be according to Resistance value i.e.,

$V_1={{{12\over7}}\over {12\over7}+6}\times12={{12\over7}\over{54\over7}}\times12={12\over54}\times12={144\over54}~V$

And,

$V_2={6\over{12\over7}+6}\times12={6\over{54\over7}}\times12={42\over54}\times12={504\over 54}~V$