Applying Kirchhoff’s Laws
Given that the equivalent resistance across CH is: \[ R_{CH} = 2R \] Equivalent Circuit Diagram
The given circuit is analyzed using Kirchhoff’s Voltage Law (KVL). We apply KVL to two loops in the circuit. Applying Kirchhoff’s Voltage Law (KVL):
In closed loop ABMNA:
Using KVL in loop ABMNA, we sum the voltage drops: \[ -3IR - 4I_1 R + 16E = 0 \]
This forms our first equation: -3IR - 4I_1 R + 16E = 0
In closed loop BCHMB:
Similarly, applying KVL in loop BCHMB: \[ -2R(I - I_1) - 6E + 4I_1 R = 0 \] This forms our second equation: -2R(I - I_1) - 6E + 4I_1 R = 0
Solving the Equations
Now, solving equations (1) and (2) simultaneously, we obtain: \[ I_1 = \frac{25E}{13R} \] Thus, the current \( I_1 \) in the circuit is determined.
A constant voltage of 50 V is maintained between the points A and B of the circuit shown in the figure. The current through the branch CD of the circuit is :
Simar, Tanvi, and Umara were partners in a firm sharing profits and losses in the ratio of 5 : 6 : 9. On 31st March, 2024, their Balance Sheet was as follows:
Liabilities | Amount (₹) | Assets | Amount (₹) |
Capitals: | Fixed Assets | 25,00,000 | |
Simar | 13,00,000 | Stock | 10,00,000 |
Tanvi | 12,00,000 | Debtors | 8,00,000 |
Umara | 14,00,000 | Cash | 7,00,000 |
General Reserve | 7,00,000 | Profit and Loss A/c | 2,00,000 |
Trade Payables | 6,00,000 | ||
Total | 52,00,000 | Total | 52,00,000 |
Umara died on 30th June, 2024. The partnership deed provided for the following on the death of a partner:
If \(\begin{vmatrix} 2x & 3 \\ x & -8 \\ \end{vmatrix} = 0\), then the value of \(x\) is: