How to Calculate Base or Emitter Current with LDR1=100Ω, LDR2=200Ω, V1=V2=1.5V

how would i calculate current flowing into base of emitters if LDR1=100ohm,LDR2=200ohms.
V1=V2=1.5V

This looks like a text book question, therefore part of course work.The first question you need to ask yourself is what sort of transistors are they, silicon or germanium, since they have different Vbe's. Assuming they are silicon, what is the voltage at the divider junction feeding the two transistor bases. Look at each transistor in turn, with both emmiters at zero volts (can you assume that, is M a current meter, and if so, what does a perfect current meter look like in terms of resistance)) and determine the conducting state of the two transistors. Knowing that, comment on the current flow.Cheers,Richard

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You can look at the R1 and R2 resistors as a voltage divider to find the voltage at the base of the two transistors. If you assume the emitter junction (which is effectively shorted to the junction between the batteries) is at zero volts, you can determine which transistor is in what mode. The base emitter voltage has to meet the +/- ~0.7V threshold to conduct current, so it would seem neither transistor meets that condition after calculating the voltage at the base to be 0.5V. Thus, the current into the base is effectively zero because the biasing scheme does not exceed the threshold needed.
Bonus: if a diode was added in the resistor voltage divider, a +/- 0.7V drop (depending on the location of the diode) would cause one of the transistors to conduct current.

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