In the circuit shown in the figure assume that the transistor is in the active region. It has a large β and its base Emitter voltage is 0.74. The value of I_c is

For the amplifier of given figure,
$${I_C}\, = \,1.3\,mA,\,{R_C}\, = \,2\,k\Omega ,\,{R_E}\, = \,500\,\Omega ,$$
$${V_T}\, = \,26\,mV,\,\beta \, = \,100,\,{V_{CC}}\, = \,15V,$$
$${V_s}\, = \,0.01\,\sin \left( {\omega t} \right)\,V\,and\,{C_b}\, = \,{C_C}\, = \,10\,\mu F.$$

(a)What is the small-signal voltage gain, $${A_V} = {V_0}/{V_s}?$$
(b)What is the approximate $${A_{v,}}\,\,if\,\,{C_e}\,\,$$ is removed?
(c)What will $${V_0}\,be\,if\,{C_b}$$ is short circuited?

The current gain of a bipolar transistor drops at high frequencies because of

If the Op-Amp in the figure, is ideal, than V₀ is