চিত্রে দেখানো ব্যাটারির emf E এবং অভ্যন্তরীণ রেজিস্ট্যান্স r হল 4.3 V এবং 1.0 Omega r...

The emf E and the internal resistance r of the battery shown in figure are 4.3 V and 1.0 Omega respectively. The external resistance R is 50 Omega . The resistances of the ammeter and voltmeter are 2.0 Omega and 200 Omega respectively. (a) Find the readings of the two meters. (b) The switch is thrown to the other side. What will be the readings of the two meters now?

In the ciruit the cells E_1 and E_2 have emfs of 4V and 8V and internal resistance 0.5 Omega and 1.0 Omega, respectively. Calculate the current through 6 Omega resistance. ltBrgt

The emf E and the internal resistance r of the battery shown in figure are 4.3V and 1.0Omega respectively. The external resistan ce R is 50Omega . The resistance of the ammeter and voltameter are 2.0Omega an d 200Omega respectively. Find the respective reading of the voltmeter.

In the circuit shown , the batteries have emf E_1 = E_2= 1V , E_3 = 2.5 V, and the resistance R_1 = 10Omega, R_2 = 20 Omega , Capacitance C = 10 muF . The charge on the left plate of the charge on the left plate of the capacitor C at steady state is

Figure shows a battery with emf 15 V in a circuit with R_1 = 30Omega, R_2 = 10 Omega, R_3 = 20 Omega and capacitance C = 10 muF . The switch S is initially in the open position and is then closed at time t= 0. What will be the fimal steady - state charge on capacitor?

A constant voltage V = 25 V is maintained between points A and B of the circuit (Fig). Find the magnitude and direaction of the current flowing through the segment CD if the resistances are equal to R_(1) = 1.0 Omega, R_(2) = 2.0 Omega, R_(3) = 3.0 Omega adn R_(4) = 4.0 Omega .

In the circuit in figure emf E_(1) = 14V (internal resistance r_(1) = 1Omega ), R_(1) = 6Omega R_(2) = 3.5 Omega, emf E_(2) = 12v (internal resistance r_(2) = 0.5Omega), C_(1) = 4muF and C_(2) = 2muF . The potential difference across R_(1) is

In the circuit in figure emf E_(1) = 14V (internal resistance r_(1) = 1Omega ), R_(1) = 6Omega R_(2) = 3.5 Omega, emf E_(2) = 12v (internal resistance r_(2) = 0.5Omega), C_(1) = 4muF and C_(2) = 2muF . The potential difference across R_(2) is

In the circuit shown in Fig. the emf of the sources is equal to xi = 5.0 V and the resistances are equal to R_(1) = 4.0 Omega and R_(2) = 6.0 Omega . The internal resistance of the source equals R = 1.10 Omega . Find the currents flowing through the resistances R_(1) and R_(2) .