Question

Please give clear answer with respective explaination , don't write in one words!!

Answer 1

Answer:

solved

Explanation:

i) in series current is the same (constant)

i = 6/3 = 2 A

power = i²R = 4x2 = 8 watts

ii) in parallel current is different (two currents i1 for 2 ohms , i2 for 1 ohm)

i1 = 4/2 = 2A

i2 = 4/1 = 4A

power = i²R = 4x2 = 8 watts

Answer 2

$\huge{\bold{\underline{\underline{....Answer....}}}}$

$\huge{\bold{\underline{Given:-}}}$

Case-1

• Potential difference = 6V.
• Resistance given are 1 ohm and 2 ohm.

Case-2

• Potential difference = 4V.
• V.Resistance given are 2 ohm and 1 ohm.

$\huge{\bold{\underline{Explanation:-}}}$

$\rule{300}{1.5}$

Finding the resultant resistance,

$\large{\bold{ \tt R_{r} = R_1 + R_2}}$

As the resistance are in series connection,

Substituting the values,

$\large{R_r = 2 + 1}$

$\large{\boxed{\tt R_r = 3 \Omega}}$

Finding the current value,

$\large{\bold{ \tt I = \dfrac{V}{R_r}}}$

Substituting,

$\large{I = \dfrac{6}{3}}$

$\large{I = \dfrac{ \cancel{6}}{ \cancel{3}}}$

$\large{\boxed{ \tt I = 2 \: A}}$

But as question says Power dissipated by 2 ohm resistance

Then,

As we know, Power formula,

$\large{\bold{ \tt P = I^2R_1}}$

Substituting the values,

$\large{P_1 = (2)^2 \times2}$

Now,

$\large{P_1 = {4 \times 2 }}$

$\large{P_1 = 2 \times 4}$

$\huge{\boxed{\boxed{ \tt P_1 = 8 \: W}}}$

$\rule{300}{1.5}$

Finding the current value,

Here the current values will not be same

$\large{\bold{ \tt I_1 = \dfrac{V}{R_1}}}$

Substituting,

$\large{I_1 = \dfrac{4}{2}}$

$\large{I_1 = \dfrac{ \cancel{4}}{ \cancel{2}}}$

$\large{\boxed{ \tt I_1 = 2 \: A}}$

As we know, Power formula,

$\large{\bold{ \tt P = I_1^2 R}}$

Substituting the values,

$\large{P_2 = \dfrac{(4)^2}{2}}$

Now,

$\large{P_2 = \dfrac{4 \times 4 }{2}}$

$\large{P_2 = \dfrac{ \cancel{4} \times 4 }{ \cancel{2}}}$

$\huge{\boxed{\boxed{ \tt P_2 = 8 \: W}}}$

Note:-

Here we have found the current that is passing through 2 ohm.

We can find the current that is passing through 1ohm.

But as the question required power of 2 ohm resistor .

So, we have founded Current flowing through 2 ohm resistor.

$\rule{300}{1.5}$

Hence from the above observation we can see that the power dissipated in second case is equal that of the first case.