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The easiest way to solve this problem ist to use Thevenin's theorem for the open terminals which arise when you cut open the circuit where the resistance $R_3$ is inserted. It is easily seen that the Thevenin voltage will be the battery voltage $V_B$ $$V_{th}= V_B$$ and the Thevenin resistance is the parallel circuit resistance of $R_1$ and $R_2$ $$R_{th}^{-1}=R_{1}^{-1}+ R_{1}^{-1}$$$$R_{th}^{-1}=R_{1}^{-1}+ R_{2}^{-1}$$ Thus the resistance of the circuit as seen from the battery is $$R=R_{th}+R_3$$ and the current flowing through the battery is $$I_B=\frac{V_B}{R}$$

The easiest way to solve this problem ist to use Thevenin's theorem for the open terminals which arise when you cut open the circuit where the resistance $R_3$ is inserted. It is easily seen that the Thevenin voltage will be the battery voltage $V_B$ $$V_{th}= V_B$$ and the Thevenin resistance is the parallel circuit resistance of $R_1$ and $R_2$ $$R_{th}^{-1}=R_{1}^{-1}+ R_{1}^{-1}$$ Thus the resistance of the circuit as seen from the battery is $$R=R_{th}+R_3$$ and the current flowing through the battery is $$I_B=\frac{V_B}{R}$$

The easiest way to solve this problem ist to use Thevenin's theorem for the open terminals which arise when you cut open the circuit where the resistance $R_3$ is inserted. It is easily seen that the Thevenin voltage will be the battery voltage $V_B$ $$V_{th}= V_B$$ and the Thevenin resistance is the parallel circuit resistance of $R_1$ and $R_2$ $$R_{th}^{-1}=R_{1}^{-1}+ R_{2}^{-1}$$ Thus the resistance of the circuit as seen from the battery is $$R=R_{th}+R_3$$ and the current flowing through the battery is $$I_B=\frac{V_B}{R}$$

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The easiest way to solve this problem ist to use Thevenin's theorem for the open terminals which arise when you cut open the circuit where the resistance $R_3$ is inserted. It is easily seen that the Thevenin voltage will be the battery voltage $V_B$ $$V_{th}= V_B$$ and the Thevenin resistance is the parallel circuit resistance of $R_1$ and $R_2$ $$R_{th}^{-1}=R_{1}^{-1}+ R_{1}^{-1}$$ Thus the resistance of the circuit as seen from the battery is $$R=R_{th}+R_3$$ and the current flowing through the battery is $$I_B=\frac{V_B}{R}$$