A circuit shown in the figure consists of a battery of emf `10V` and two capacitance `C_(1)` and `C_(2)` of capacitances `1.0muF` and `2.0muF` respectively. The potential difference `V_(A) - V_(B)` is `5V`

Two capacitors C_(1) and C_(2) are connected in a circuit as shown in figure. The potential difference (V_(A) - V_(B)) is

A circuit has section AB as shown in figure. The emf of the source equals E=10V , the capacitor capacitances are equal to C_1 =1.0muF and C_2=2.0muF , and the potential difference V_A-V_B=5.0V . Find the voltage across each capacitor.

A circuit has section AB as shown in figure. The emf of the source equals E=10V , the capacitor capacitances are equal to C_1 =1.0muF and C_2=2.0muF , and the potential difference V_A-V_B=5.0V . Find the voltage across each capacitor.

A circuit has a section Ab shown in fig. The emf of the source equals E = 10V , the capacitances are equal to C_(1) = 1.0 muF and C_(2) = 2.0 muF , and the potential difference varphi_(A) - varphi_(B) = 5.0 V . Find the voltage across each capacitor.

A circuit has a section AB as shown in Fig. HQ 7. The emf of the source equals E = 10V, the capacitor capacitances are equal to C_1 = 1.0 muF and C_2 = 2.0 muF and the potential differences V_A - V_B = 5.0 V . Find the voltage across each capacitor.

A circuit has section AB as shown in The emf of the cell is 10 V , and the capacitors have capacitances C_(1)=1 muF and C_(2)=2 muF . .

In each capacitance C_(1) is 6.0 muF , and each capacitance C_(2) is 4.0 muF . . With 420 V across a and b the value of V_(c)-V_(d) is.

In each capacitance C_(1) is 6.0 muF , and each capacitance C_(2) is 4.0 muF . . With 420 V across a and b the value of V_(c)-V_(d) is.

Three capacitors of capacitance 1.0muF,2.0muF" and "5.0muF are connected in series to a 10V source. The potential difference across the 2.0muF capacitor is

In each capacitance C_(1) is 6.0 muF , and each capacitance C_(2) is 4.0 muF . . The charge on C_(1) nearest to a when V_(ab)=420 V is.