In Fig. 27-41,${\mathbf{R}}_{\mathbf{1}}\mathbf{=}\mathbf{10}\mathbf{\Omega }$,${\mathbf{R}}_{\mathbf{2}}\mathbf{=}\mathbf{20}\mathbf{\Omega }$, and the ideal batteries have emfs ${\mathbf{\epsilon }}_{\mathbf{1}}\mathbf{=}\mathbf{20}\mathbf{.}\mathbf{0}\mathbf{V}$and ${\mathbf{\epsilon }}_{\mathbf{2}}\mathbf{=}\mathbf{50}\mathbf{.}\mathbf{0}\mathbf{V}$. What value of results in no current through battery 1?

• a) The given resistance values are$R_1=10\Omega$,$R_2=20\Omega$.
• b) Emf of the ideal batteries is${\epsilon }_1=20.0V,{\epsilon }_2=50.0V$.

Kirchhoff's voltage law states that the total voltage across a closed loop is zero. The voltage here is given by the names of the ideal batteries connected with the voltage using Ohm's law for the given resistors of the circuit. Similarly, the current junction rule defines that the current going inside the circuit is equal to the net current coming out of it.

Formulae:

The voltage equation using Ohm’s law,

V=IR (i)

Kirchhoff’s voltage law,

$\sum _{closedloop}V=0$ (ii)

Kirchhoff’s junction rule,

I_in = I_out (iii)

Here R is the resistance, I is the current.

Using equation (i) in the loop rule of equation (ii), the voltage equation for the left loop can be given as follows:

$$\begin{gathered} {\epsilon }_1-i_1{R}_1-i_3{R}_3=0 \\ 20.0V-i_1{R}_1-I_3{R}_3=0 \end{gathered}$$

Now, the voltage equation for the right loop can be given using equation (i) in equation (ii) as follows:

$$\begin{gathered} {\epsilon }_1-i_1{R}_1-i_2{R}_2-{\epsilon }_2=0 \\ 20.0V-i_1{R}_1-i_2{R}_2-50.0V=0 \end{gathered}$$

Now, using the junction rule of equation (iii), the current equation can be given as follows:

i₂+ i₃=i₁ (c)

Requiring no current through battery 1 means that i₁.=0.

Now, using this value in equation (c), the current through resistors 2 and 3 can be given as:

i₂ = -i₃

Now, equation(a) becomes:
$$\begin{gathered} 20.0V-i_3{R}_3=0 \\ {i}_3=\frac{20.0V}{R_3} \end{gathered}$$

And using the above values in equation (b), the resistance value of resistor 3 can be given as follows:

$$\begin{gathered} -30.0V-\left(-i_3\right)R_2=0 \\ -30.0V+\left(\frac{20.0V}{R_3}\right)R_2=0 \end{gathered}$$

role="math" localid="1662223582939" $$\begin{gathered} R_3=\frac{20.0V}{30.0V}\left(20.0\Omega \right) \\ =\frac{40}{3}\Omega \\ =13.3\Omega \end{gathered}$$

Hence, the value of the resistance is$13.3\Omega$ .