A) \[{{V}_{B}}=40mV,{{V}_{w}}=135mV\] with left side of pilot at higher voltage.
B) \[{{V}_{B}}=40mV,{{V}_{w}}=135mV\] with right side of pilot at high voltage.
C) \[{{V}_{B}}=45mV,{{V}_{w}}=120mV\] with left side of pilot at higher voltage.
D) \[{{V}_{B}}=45mV,{{V}_{w}}=120mV\]with right side of pilot at higher voltage.
Correct Answer: C
Solution :
\[{{V}_{B}}={{B}_{4}}\left( 5 \right)\left( 240 \right)\] \[{{B}_{H}}=B\cos \theta \] \[{{B}_{H}}\frac{5\sqrt{5}\times {{10}^{-5}}}{3}\] \[{{B}_{v}}=\frac{10}{3}\times {{10}^{-5}}T\] \[{{V}_{ & B}}=\frac{5\sqrt{5}}{3}\times {{10}^{-5}}\times 5\times 240\] \[{{V}_{B}}=44.6mV=45mV\] \[{{V}_{w}}=Bv\ell V\] \[={{10}^{-4}}\times 1200\] \[{{V}_{\omega }}=120mV\] (left side at fighter voltage)You need to login to perform this action.
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