The current in the ammeter becomes 1/e of the initial value after 1 second
The current in the ammeter becomes zero after a long time.
The voltmeter displays −5V as soon as the key is pressed, and displays +5V after a long time
The voltmeter will display 0 V at time t=ln2 seconds
Writing instantaneous equation, with q1 being charge on upper and q2 being that on lower capacitor, q1=(200×10⁻⁷)(1−e⁻ᵗ) q2=(100×10⁻⁷)(1−e⁻ᵗ) q1/C=(50×10³)(dq2/dt) 200×10⁻⁷(1−e⁻ᵗ)/40×10⁻⁶=(50×10³)(100×10⁻⁷)e⁻ᵗ e⁻ᵗ=0.5 t=ln(2) I=I1+I2=(200×10⁻⁷)(e⁻ᵗ)+(100×10⁻⁷)e⁻ᵗ=100×10⁻⁷[2e⁻ᵗ+e⁻ᵗ]=0.3e⁻ᵗ At t=∞,I=0 Answer is A, B, C and D.