1: Analyzing the Circuit
The given circuit consists of resistors in series and parallel. First, simplify the resistances step by step. 1. Combine the 10\(\mu\) and 10\(\mu\) resistors in parallel between points B and C: \[ R_{\text{BC}} = \frac{1}{\frac{1}{10} + \frac{1}{10}} = 5 \] 2. Combine the 20\(\mu\) resistors in parallel between points C and D: \[ R_{\text{CD}} = \frac{1}{\frac{1}{20} + \frac{1}{20}} = 10 \] 3. Simplify the overall network of resistors to find the effective resistance between points A and M: \[ R_{\text{eff}} = 12 \]
2: Power Supplied by the Battery
The power supplied by the battery is given by: \[ P = \frac{V^2}{R_{\text{eff}}} = \frac{6^2}{12} = 3W \] Thus, the power supplied by the battery is \( P = 3W \).
Two cells of emf 1V and 2V and internal resistance 2 \( \Omega \) and 1 \( \Omega \), respectively, are connected in series with an external resistance of 6 \( \Omega \). The total current in the circuit is \( I_1 \). Now the same two cells in parallel configuration are connected to the same external resistance. In this case, the total current drawn is \( I_2 \). The value of \( \left( \frac{I_1}{I_2} \right) \) is \( \frac{x}{3} \). The value of x is 1cm.
Current passing through a wire as function of time is given as $I(t)=0.02 \mathrm{t}+0.01 \mathrm{~A}$. The charge that will flow through the wire from $t=1 \mathrm{~s}$ to $\mathrm{t}=2 \mathrm{~s}$ is:
The correct IUPAC name of \([ \text{Pt}(\text{NH}_3)_2\text{Cl}_2 ]^{2+} \) is: