Question

A network of 4 ohm and 3 ohm resistance are connected to a 10 volt battery. Compute the equivalence resistance and current obtained from battery ​

Answer 1

Answer:

1)If they are connected in series

the equivalent resistance is

R1+R2=4+3=7ohm

the current is v/r=10/7=1.428~1.43A

2)If the resistance are connected in parallel

equivalent resistance is equal to

1/4+1/3=7/12=0.58333~0.58ohm

the current is 10/0.58=17.2413~17.24

Answer 2

$\sf \underline{ \fcolorbox{red}{pink}{ \huge{Solution :)}}}$

Given ,

• R1 = 4 ohm
• R2 = 3 ohm
• Potential difference (V) = 10 volt

Case I : if resistance are connected in a series

We know that , for series combination

$\large \sf \fbox{Resistance = R_{1} + R_{2} }$

Substitute the known values , we get

R = 4 + 3

R = 7 ohm

We know that ,

$\large \sf \fbox{V = R \times I}$

Substitute the known values , we get

➡I = 10/7

➡I = 1.4 Amp

Hence , in series combination, the equivalence resistance and current obtained from battery are 7 ohm and 1.4 amp

Case II : if resistance are connected in a parallel

We know that , for parallel combination

$\large \sf \fbox{\frac{1}{R} = \frac{1}{ R_{1}} + \frac{1}{ R_{2}}}$

Substitute the known values , we get

1/R = 1/4 + 1/3

1/R = (3 + 4)/12

1/R = 7/12

R = 12/7

R = 1.7

We know that ,

$\large \sf \fbox{V = R \times I}$

Substitute the known values , we get

I = 10/1.7

I = 100/17

I = 5.8

Hence , in parallel combination, the equivalence resistance and current obtained from battery are 1.7 ohm and 5.8 amp

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