Understanding KVL Loop in BJT Biasing and Calculating Ve

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karandeepdps
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thread moved into homework forum, so comes without template
Circuit schematic of an NPN transistor with base resistor RB to ground, collector resistor RC to +VCC, emitter resistor RE to -VEE, and voltage nodes marked VB, VE and VC, with grounds shown.
Circuit schematic of an NPN transistor with base resistor RB to ground, collector resistor RC to +VCC, emitter resistor RE to -VEE, and voltage nodes marked VB, VE and VC, with grounds shown.
Solution text: (i) The emitter voltage w.r.t. ground is V_E = -V_EE + I_E R_E = -10V + 1.8 mA × 4.7 kΩ = -1.54V, with the result -1.54V in red.

Please help how Ve is calculated what i think is that it should be:
-Ve=-Vee+IeRe;
How Ve is taken positive please someone draw equivalent circuit for same
 
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VE is defined as the emitter voltage wrt ground. Nothing about this says that VE must turn out to be a positive voltage, though. You can see that VE will be somewhere between -VEE and +VCC.

If NPN emitter current is the current out of the transistor emitter, then the upper end of RE is the more positive. Draw arrows to show current direction, and voltage drop across resistances, and apply Kirchoff's Voltage Law.
 
Thanks will you please help me in this question also i am confused in Vcc equations
Textbook solution and circuit schematic of a CE transistor amplifier with base resistor biasing, Fig. 9.10, showing Vcc = 15 V, RC and RB resistors, and the KVL equations Vcc = VCE + IC RC and Vcc = IB RB + VBE with resulting values.
 
Explain which line in the solution you are having trouble with.
 
Last line,how they have applied kvl to an open loop.
 
karandeepdps said:
Last line,how they have applied kvl to an open loop.
There is no open loop.

RB is supplying base current from the positive rail, VCC. Because the transistor is operating in the active region, its base is forward biased and its B-E potential is labelled VBE.

The wire from the base ending in mid-air simply indicates where the amplifier's AC input will be impressed when the circuit is later completed to function as an audio amplifier, for example.