Question

In the circuit shown all five resistors have the same value $200\Omega$ and each cell has an emf $3volts.$ Find the open circuit voltage and the short circuit current for the terminals $A$ and $B.$

Answer 1

$R_{TH}=\frac{600\times 400}{600+400}\Omega$

$=\frac{2400}{10}=240\Omega$

So, in the loop we have,

$5\times 200I=3+3-3$

$⟹I=\frac{3}{1000}A=3mA$

$\therefore V_A-V_B=0.2\times 3+3+0.2\times 3$

$=3+1.2=4.2V-V_{TH}$

So, $I_n$ or short circuit current is,

$I_n=\frac{V_{TH}}{R_{TH}}=\frac{4.2V}{240\Omega }A=17.5mA$