Name: Kirchhoff's Second Law
Uploaded: 2019-07-19
Duration: 5 min 21 s
Description: Kirchhoff’s first law is very handy when calculating currents. Here, we explore his second potent law and use it to reduce complicated loops. Watch and learn.

Question

Two resistors are connected in series with

5 V

battery of negligible internal resistance. A current of

2 A

flows through each resistor, if they are connected in parallel with the same battery a current of

\frac{25}{3} A

flows through combination. Calculate the value of each resistance.

Accepted Answer

Given, battery of potential difference: $E = 5 V$

When the resistors are connected in series then the net resistance of resistors is series will be: $R_{S} = \frac{E}{I} = \frac{5 V}{2 A} = 2.5 Ω$

Also, the equivalent series resistance will be: $R_{S} = R_{1} + R_{2}$ ...................... $(1)$

When the resistors are connected in parallel then the net resistance of resistors is series will be: $R_{P} = \frac{E}{I} = \frac{5 V}{\frac{25}{3} A} = 0.6 Ω$

Also, the equivalent parallel resistance will be: $R_{P} = \frac{R_{1} R_{2}}{R_{1} + R_{2}}$ ...................... $(2)$

By substituting equation $(1)$ in equation $(2)$ we get, $R_{P} = \frac{R_{1} R_{2}}{R_{S}} \Rightarrow R_{1} R_{2} = R_{P} R_{S} = 1.5 Ω$

Now, ${(R_{1} - R_{2})}^{2} = {(R_{1} + R_{2})}^{2} - 4 R_{1} R_{2}$

By, substituting values we get ${(R_{1} - R_{2})}^{2} = {(2.5 Ω)}^{2} - 4 (1.5 Ω) = 0.25 Ω$ ...................... $(3)$

$R_{1} - R_{2} = 0.5 Ω$ ...................... $(4)$

By adding equation $(1)$ and $(4)$ , we get $2 R_{1} = 3 Ω \Rightarrow R_{1} = 1.5 Ω$

Now, substituting $R_{1}$ into $(4)$ we get, $R_{2} = 1 Ω$

Hence, the value of the two resistances are $1.5 Ω$ and $1 Ω$ .

Two resistors are connected in series with $5 V$ battery of negligible internal resistance. A current of $2 A$ flows through each resistor, if they are connected in parallel with the same battery a current of $\frac{25}{3} A$ flows through combination. Calculate the value of each resistance.

Important Questions on Current Electricity

Learn and Prepare for any exam you want !

Two resistors are connected in series with 5 V battery of negligible internal resistance. A current of 2 A flows through each resistor, if they are connected in parallel with the same battery a current of 253 A flows through combination. Calculate the value of each resistance.

Important Questions on Current Electricity

Learn and Prepare for any exam you want !

Two resistors are connected in series with $5 V$ battery of negligible internal resistance. A current of $2 A$ flows through each resistor, if they are connected in parallel with the same battery a current of $\frac{25}{3} A$ flows through combination. Calculate the value of each resistance.