Question

find the potential of J with respect of G​

Answer 1

Answer:

20 V

Let the current in wire be i .

Use kvl

$60 - 64i - 32i = 0 \\ \\ 96i = 60 \\ \\ i = \frac{60}{96} = \frac{5}{8} \: amp$

Now voltage across ends of 32 ohm resistance .

$v = ir = \frac{5}{8} \times 32 = 20 \: V$

So when we go G to J there comes voltage difference of 20 Volts .

Answer 2

Given that,

Potential = 60 V

Resistance $R_{1}=64\ \Omega$

Resistance $R_{2}=32\ \Omega$

R₁ and R₂ is connected in series.

We need to calculate the current

Using ohm's law

$I=\dfrac{V}{R_{1}+R_{2}}$

Put the value into the formula

$I=\dfrac{60}{64+32}$

$I=\dfrac{5}{8}\ A$

We need to calculate the potential at J

Using ohm's law

$V_{J}= IR$

$V_{J}=60-64\times\dfrac{5}{8}$

$V_{J}=20\ V$

According to figure,

$V_{G}$ is zero because it is grounded.

We need to calculate the potential of J with respect of G​

Using formula of relative potential

$V_{JG}=V_{J}-V_{G}$

Put the value into the formula

$V_{JG}=20-0$

$V_{JG}=20\ V$

Hence, The potential of J with respect of G​ is 20 V.