Kirchhoff's Laws PDF

## 10.2 Kirchhoff's Laws **List the known quantities:** * $V_1 = 0.5 V$ * $V_2 = 1 V$ * $V_3 = 2 V$ * Resistance of each resistor, $R = 0.25 Ω$ **Label the diagram to show the loops, polarity of the batteries and directions of each current:** **A diagram that would score 1 mark should look lik...

## 10.2 Kirchhoff's Laws **List the known quantities:** * $V_1 = 0.5 V$ * $V_2 = 1 V$ * $V_3 = 2 V$ * Resistance of each resistor, $R = 0.25 Ω$ **Label the diagram to show the loops, polarity of the batteries and directions of each current:** **A diagram that would score 1 mark should look like this:** A diagram with a circuit containing 4 resistors ($R_1$, $R_2$, $R_3$ and $R_4$) in series with each other. The positive terminal of each battery ($V_1$, $V_2$ and $V_3$) is connected to a resistor in series with a wire going to the next resistor. The circuit is then connected back to the negative terminal of the battery. The current is labelled $I_1$ for the first loop and is shown entering the circuit via $V_1$. The current is labelled $I_3$ in the next loop and is shown entering the circuit via $V_3$. Finally, the current $I_2$ is shown in the next loop going from $V_3$ to $V_2$. * Currents $I_1$, $I_2$ and $I_3$ are labelled as shown with arrows in the correct direction * Positive and negative battery and resistor terminals should be indicated correctly with the positive side of the resistor the side where the current enters; [1 mark] **Use Kirchhoff's first law to obtain an equation for the currents at junction A:** * $I_1 + I_3 = I_2$ [1 mark] **Use Kirchhoff's second law to obtain an equation for the sum of the potential differences in loops 1 and 2:** * Recall Ohm's Law: $V = IR$ ###### Loop 1: * $V_1 - I_1R_1 - I_2R_2 - V_2 - I_1R_4 = 0$ * $0.5 - (0.25I_1) - (0.25I_2) - 1 - (0.25I_1) = 0$ * $-0.5 - 0.5I_1 - 0.25I_2 = 0$ * $0.5I_1 + 0.25I_2 = -0.5$ ###### AND ###### Loop 2: * $V_3 - I_3R_3 - I_2R_2 - V_2 = 0$ * $2 - (0.25I_3) - (0.25I_2) - 1 = 0$ * $ 1 - (0.25 I_3) - (0.25I_2) = 0$ * $1 = 0.25I_3 + 0.25I_2$ * $I_3 + I_2 = 4$ [1 mark] **Substitute for $I_3$ into the equation for current:** * From loop 2: $I_3 = 4 - I_2$ * Equation for current: $I_1 + I_3 = I_2$ * So, $I_1 + (4 - I_2) = I_2$ * $I_1 = 2I_2 - 4$ [1 mark] **Now substitute for $I_1$ from above into the equation for loop 1:** * From loop 1: $0.5I_1 + 0.25I_2 = -0.5$ * $0.5 (2I_2 - 4) + 0.25I_2 = -0.5$ * $I_2 - 2 + 0.25I_2 = -0.5$ * $1.25 I_2 = -0.5 + 2$ * $I_2 = 1.2 A$ [1 mark] **Now substitute for $I_2$ to obtain $I_1$:** * $I_1 = (2 \times 1.2) - 4$ * $I_1 = -1.6 A$ [1 mark] **Substitute for $I_2$ into the equation for loop 2:** * From loop 2: $I_3 + I_2 = 4$ * $I_3 = 4 - 1.2$ * $I_3 = 2.8 A$ [1 mark] **[Total: 7 marks]** You can label your circuit in any way but make sure the direction of the current is consistent all the way around. When obtaining an equation for the sum of the potential differences if you cross from: * terminal of one resistor to the + terminal of another resistor → the change in potential difference is + * + terminal of one resistor to the - terminal of another resistor → the change in potential difference is - It is always a good idea to check your answer at the end. Here you can do this by using the equation for current: * $I_1 + I_3 = I_2$ * So, $-1.6 + 2.8 = 1.2$ which is correct