The potential difference between points C and D of the electrical circuit shown in the figure is
Show Hint
When solving circuit problems in competitive exams, if your calculated answer doesn't match any option, quickly recheck your calculations. If the calculation is correct, suspect a typo in the question's data. You can sometimes work backwards from the given options to identify the likely typo, as shown here.
Step 1: Understanding the Concept:
To find the potential difference between points C and D ($V_C - V_D$), we need to determine the current flowing through the resistor connecting these two points. We can determine this current by applying Kirchhoff's Current Law (KCL) sequentially at nodes B and C. According to KCL, the algebraic sum of currents entering a node is equal to the sum of currents leaving it.
Step 2: Analyzing Node B:
First, we determine the current flowing from B to C ($I_{BC}$).
Current entering node B from A ($I_{AB}$) = $4 \, \text{A}$.
Current entering node B from E ($I_{EB}$) = $1.8 \, \text{A}$ (The arrow on branch EB points downwards towards B).
Current leaving node B towards C is $I_{BC}$.
Applying KCL at Node B:
\[ \sum I_{\text{in}} = \sum I_{\text{out}} \]
\[ I_{AB} + I_{EB} = I_{BC} \]
\[ 4 + 1.8 = I_{BC} \]
\[ I_{BC} = 5.8 \, \text{A} \]
Step 3: Analyzing Node C:
Next, we determine the current flowing from C to D ($I_{CD}$).
Current entering node C from B ($I_{BC}$) = $5.8 \, \text{A}$.
Current leaving node C towards F ($I_{FC}$) = $1.3 \, \text{A}$ (The arrow on branch FC points upwards, away from C).
Current leaving node C towards G ($I_{CG}$) = $1 \, \text{A}$ (The arrow on branch CG points downwards, away from C).
Current leaving node C towards D is $I_{CD}$.
Applying KCL at Node C:
\[ I_{BC} = I_{FC} + I_{CG} + I_{CD} \]
\[ 5.8 = 1.3 + 1 + I_{CD} \]
\[ 5.8 = 2.3 + I_{CD} \]
\[ I_{CD} = 5.8 - 2.3 \]
\[ I_{CD} = 3.5 \, \text{A} \]
Step 4: Calculation of Potential Difference:
Now that we have the current $I_{CD}$ and the resistance $R_{CD} = 8 \, \Omega$, we can calculate the potential difference using Ohm's Law ($V = IR$).
\[ V_{C} - V_{D} = I_{CD} \times R_{CD} \]
\[ V_{C} - V_{D} = 3.5 \, \text{A} \times 8 \, \Omega \]
\[ V_{C} - V_{D} = 28 \, \text{V} \]
Final Answer:
The potential difference between points C and D is 28 V.
